IISER Mathematics - Vector Algebra
Exam Duration: 45 Mins Total Questions : 30
Let P,Q,R be three points with respective position vectors i+j,i-j and ai+bj+ck.The points P,Q,R are collinear, if
- (a)
a = b = c = 1
- (b)
a = b = c = 0
- (c)
a = 1,b and c arbitrary scalars
- (d)
a and b are arbitrary scalars and c=0
The three given vectors can be collinear if these are coplanar, for which
\(\left| \begin{matrix} 1 & 1 & 0 \\ 1 & -1 & 0 \\ a & b & c \end{matrix} \right| =0\)
⇒ -2c = 0
⇒ c = 0, whatever a and b may be.
If \(\vec { a } \quad =\quad i+2j+3k,\quad and\quad \vec { b } \quad =\quad 3i+6j+2k,\) the vector in the direction of \(\vec { a } \) and having magnitude \(\left| \vec { b } \right| \) is
- (a)
7(i+2j+2k)
- (b)
\(\frac { 7 }{ 9 } (i+2j+2k)\)
- (c)
\(\frac { 7 }{ 3 } (i+2j+2k)\)
- (d)
NONE OF THESE
The vector \(\overrightarrow{c}\) in the direction of the vector \(\overrightarrow{a}\) having magnitude \(\overrightarrow{b}\) is
\(\overrightarrow{c}\) = |\(\overrightarrow{b}\)|\(\overrightarrow{a}\over|\overrightarrow{a}|\)
Now, |\(\overrightarrow{b}\)| = \(\sqrt{3^2+6^2+2^2}=7\)
|\(\overrightarrow{a}\)| = \(\sqrt{1^2+2^2+2^2}=3\)
ஃ \(\overrightarrow{c}\)=\(7\over3\)(i+2j+2k)
The value of s for which the points with position vectors -(j+k), 4i+5j+sk, 3i+9j+4k and 4(-i+j+k) are complanar is
- (a)
-1
- (b)
0
- (c)
1
- (d)
NONE OF THESE
Let the given points be A, B, C, D respectively AB = (4i + 5j + sk) - 4i + 6j + (s+1)k
Similarly
\(\overrightarrow{AC}\) = 3i + 10j + 5k, \(\overrightarrow{AD}\) = -4i + 5j + 5k
The given points are coplanar, if
\(\overrightarrow{AB},\overrightarrow{AC}\overrightarrow{AD}\) are coplanar, for which
\(\overrightarrow{AB}.(\overrightarrow{AC}\times\overrightarrow{AD}) \) are coplanar, for which
\(\overrightarrow{AB}.(\overrightarrow{AC}\times\overrightarrow{AD})\) = \(\left| \begin{matrix} 4 & 6 & s+1 \\ 3 & 10 & 5 \\ -4 & 5 & 5 \end{matrix} \right| =0\)
⇒ 55(s-1) = 0
⇒ s = 1.
If \(\vec { a } \) is non-zero vector, and \(\left| k\vec { a } \right| =1,\) the value of k is
- (a)
\(\frac { 1 }{ \vec { a } } \)
- (b)
\(\frac { 1 }{ \left| \vec { a } \right| } \)
- (c)
\(\pm \frac { 1 }{ \left| \vec { a } \right| } \)
- (d)
NONE OF THESE
Since, \(|\overrightarrow{ka}|=1 \)
or \(\pm k|\overrightarrow{a}|=1\)
or k = \(\pm{1\over|\overrightarrow{a}|}\)
If \(\vec { a } \quad =\quad 2i+3j-k,\quad \vec { b } \quad =\quad -i+2j-4k,\quad \vec { c } \quad =\quad i+j+k\) then value of \({ (\vec { a } \times \vec { b } ) }.{ (\vec { a } \times \vec { c } ) }\) is
- (a)
60
- (b)
64
- (c)
74
- (d)
-74
\(\overrightarrow{a}\times\overrightarrow{b}\) = \(\left| \begin{matrix} i & j & k \\ 2 & 3 & -1 \\ -1 & 2 & -4 \end{matrix} \right| \) = -10i + 9j +7k
\(\overrightarrow{a}\times\overrightarrow{c}\) = \(\left| \begin{matrix} i & j & k \\ 2 & 3 & -1 \\ 1 & 1 & 1 \end{matrix} \right| \)= 4i-3j-k
∴ \(\overrightarrow{a}\times\overrightarrow{b}\).\(\overrightarrow{a}\times\overrightarrow{c}\)
= (-10i + 9j + 7k).(4i-3j-k)
= -40 - 27 - 7 = -74
If \(\vec { a } ,\vec { b } ,\vec { c } \) are unit vectors, then \({ \left| \vec { a } -\vec { b } \right| }^{ 2 }+{ \left| \vec { b } -\vec { c } \right| }^{ 2 }+{ \left| \vec { c } -\vec { a } \right| }^{ 2 }\), does not exceed
- (a)
4
- (b)
9
- (c)
8
- (d)
6
If \(a=\lambda \hat { i } +2\hat { j } -3\hat { k } ,b=2\hat { i } +\lambda \hat { j } -\hat { k } ,c=\hat { i } +2\hat { j } +\hat { k } \) and [a b c]=6, then \(\lambda \) is equal to
- (a)
-8 or 3
- (b)
-9 or 3
- (c)
-3 or +9
- (d)
8 or 5
[a b c] = \(\left[ \begin{matrix} \lambda & 2 & -3 \\ 2 & \lambda & -1 \\ 1 & 2 & 1 \end{matrix} \right] =6\)
On applying,\(C_1\rightarrow C_1-C_3\) we get
6 = \(\lambda^2+5\lambda-18\)
\(\Rightarrow \quad { \lambda }^{ 2 }+5{ \lambda }-24=0\)
\(\therefore \quad \quad { \lambda }=-8\quad or\quad 3\)
The three vectors \(a=\hat { i } -\hat { k } ,b=\alpha \hat { i } +\hat { j } +(1-\alpha )\hat { k } and\quad c=\beta \hat { i } +\alpha \hat { j } +(1+\alpha -\beta )\hat { k } \) . Then, [a b c] depends on
- (a)
\(only\quad \alpha \)
- (b)
\(only\quad \beta \)
- (c)
\(both\quad \alpha \quad and\quad \beta \)
- (d)
\(Neither\quad \alpha \quad nor\quad \beta \)\(\)
\([a\quad b\quad c]=\left| \begin{matrix} 1 & 0 & -1 \\ \alpha & 1 & 1-\alpha \\ \beta & \alpha & 1+\alpha -\beta \end{matrix} \right| \)
\([a\quad b\quad c]=\left| \begin{matrix} 1 & 0 & -1 \\ \alpha & 1 & 1-\alpha \\ \beta & \alpha & 1+\alpha -\beta \end{matrix} \right| \) [using \(C_3\rightarrow C_3+C_1\)]
So, [a b c] depends on neither \(\alpha\ nor\ \beta\).
If a,b,c are non - coplanar vectors and \(\lambda \) is a real number, then \(\left[ \lambda (a+b){ \lambda }^{ 2 }b\lambda c \right] =[a(b+c)b]\) for
- (a)
exactly two values of \(\lambda \)
- (b)
exactly three values of \(\lambda \)
- (c)
no value of \(\lambda \)
- (d)
exactly one value of \(\lambda \)
Given [\(\lambda\)(a+b)\(\lambda\)2b\(\lambda\)c] = [a b + c b]
Let \(a=a_1\widehat{i}+a_2\widehat{j}+a_3\widehat{k},b=b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}\)
and \(c=c_1\widehat{i}+c_2\widehat{j}+c_3\widehat{k}\)
ஃ \(\left| \begin{matrix} \lambda ({ a }_{ 1 }+{ b }_{ 1 }) & \lambda ({ a }_{ 2 }+{ b }_{ 2 }) & \lambda ({ a }_{ 2 }+{ b }_{ 2 }) \\ { \lambda }^{ 2 }{ b }_{ 1 } & { \lambda }^{ 2 }{ b }_{ 2 } & { \lambda }^{ 2 }{ b }_{ 3 } \\ { \lambda c }_{ 1 } & { \lambda c }_{ 2 } & { \lambda c }_{ 3 } \end{matrix} \right| =\left| \begin{matrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 }+{ c }_{ 1 } & { b }_{ 2 }+{ c }_{ 2 } & { b }_{ 3 }+{ c }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \end{matrix} \right| \)
⇒ \({ \lambda }^{ 4 }\left| \begin{matrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =-\left| \begin{matrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| \) [by property of determinant]
⇒ \({ \lambda }^{ 4 }=-1\)
So, no real value of \(\lambda\) exists
a,b,c are three vectors, such that a+b+c=0, |a|=1, |b|=2, |c|=3, then a.b + b.c + c.a is equal to
- (a)
0
- (b)
-7
- (c)
7
- (d)
1
Given, |a|=1, |b|=2, |c|=3 and a+b+c=0
Now, (a+b+c)2 =|a|2 + |b|2 + |c|2 + 2(a.b+b.c+c.a)
0 = 12 + 22 + 32 +2(a.b+b.c+c.a)
2(a.b+b.c+c.a) = -14
a.b + b.c + c.a = -7
If \(\begin{vmatrix} a & { a }^{ 2 } & 1+{ a }^{ 3 } \\ b & { b }^{ 2 } & 1+{ b }^{ 3 } \\ c & { c }^{ 2 } & 1+{ c }^{ 3 } \end{vmatrix}=0\) and vectors (1,a,a2), (1,b,b2) and (1,c,c2) are non - coplanar, then the product abc equals
- (a)
2
- (b)
-1
- (c)
1
- (d)
0
Given, \(\left| \begin{matrix} a & { a }^{ 2 } & 1+{ a }^{ 3 } \\ b & { b }^{ 2 } & 1+{ b }^{ 3 } \\ c & { c }^{ 2 } & 1+{ c }^{ 3 } \end{matrix} \right| =\left| \begin{matrix} a & { a }^{ 2 } & 1 \\ b & { b }^{ 2 } & 1 \\ c & { c }^{ 2 } & 1 \end{matrix} \right| +\left| \begin{matrix} a & { a }^{ 2 } & { a }^{ 3 } \\ b & { b }^{ 2 } & { b }^{ 3 } \\ c & { c }^{ 2 } & { c }^{ 3 } \end{matrix} \right| \)=0
\(\Rightarrow \left| \begin{matrix} a & { a }^{ 2 } & 1 \\ b & { b }^{ 2 } & 1 \\ c & { c }^{ 2 } & 1 \end{matrix} \right| +abc\left| \begin{matrix} a & { a }^{ 2 } & 1 \\ b & { b }^{ 2 } & 1 \\ c & { c }^{ 2 } & 1 \end{matrix} \right| =0\)
\(\Rightarrow (1+aabc)\left| \begin{matrix} a & { a }^{ 2 } & 1 \\ b & { b }^{ 2 } & 1 \\ c & { c }^{ 2 } & 1 \end{matrix} \right| =0\quad \left[ \therefore \left| \begin{matrix} a & { a }^{ 2 } & 1 \\ b & { b }^{ 2 } & 1 \\ c & { c }^{ 2 } & 1 \end{matrix} \right| \neq 0 \right] \)
\(\Rightarrow \quad 1+abc=0\Rightarrow abc=-1\)
The vector \(\hat { i } +x\hat { j } +3\hat { k } \) is rotated through an angle \(\theta \) and doubled in magnitude, then it becomes \(4\hat { i } +(4x-2)\hat { j } +2\hat { k } .\) The values of x are
- (a)
\(\left\{ -\frac { 2 }{ 3 } ,2 \right\} \)
- (b)
\(\left( \frac { 1 }{ 3 } ,2 \right) \)
- (c)
\(\left( \frac { 2 }{ 3 } ,0 \right) \)
- (d)
\(\left\{ 2,7 \right\} \)
Since, the vector \(\widehat{i}+x\widehat{j}+3\widehat{k}\) is doubled in magnitude, then it becomes
\(4\widehat{i}+(4x-2)\widehat{j}+2\widehat{k}\)
ஃ \(2|\widehat{i}+x\widehat{j}+3\widehat{k}|=|4\widehat{i}+(4x-2)\widehat{j}+2\widehat{k|}\)
⇒ \(2\sqrt{1+x^2+9}=\sqrt{16+(4x-2)^2+4}\)
⇒ 40+4x2 = 20+(4x-2)2
⇒ 3x2 - 4x - 4 = 0
⇒ (x-2)(3x+2) = 0
ஃ x = 2, -\(2\over3\)
Let \(\vec { a } \),\(\vec { b } \),\(\vec { c } \) be three unit vectors such that 3 \(\vec { a } \) + 4 \(\vec { b } \) + 5 \(\vec { c } \) = 0. Then which of the following statements is true?
- (a)
\(\vec { a } is parallel to \vec b\)
- (b)
\(\vec { a } is perpendicular to \vec b\)
- (c)
\(\vec { a } is neither parallel nor perpendicular to \vec b\)
- (d)
none of the above
3 \(\vec { a } \) + 4 \(\vec { b } \) + 5 \(\vec { c } \) = 0
\(\Rightarrow\) \(\vec { a } \), \(\vec { b } \), \(\vec { c } \) are coplanar
No other conclusion can be derived from it.
If \(\left| \vec { a } +\vec { b } \right| =\left| \vec { a } -\vec { b } \right| \), then
- (a)
\(\vec a\) is parallel to \(\vec b\)
- (b)
\(\vec { a } \bot \vec { b } \)
- (c)
\(\left| \vec { a } \right| =\left| \vec { b } \right| \)
- (d)
none of these
\(\left| \vec { a } +\vec { b } \right| =\left| \vec { a } -\vec { b } \right| \\ \Rightarrow { \left| \vec { a } +\vec { b } \right| }^{ 2 }={ \left| \vec { a } -\vec { b } \right| }^{ 2 }\\ \Rightarrow { \left| \vec { a } \right| }^{ 2 }+{ \left| \vec { b } \right| }^{ 2 }+2\vec { a } .\vec { b } ={ \left| \vec { a } \right| }^{ 2 }+{ \left| \vec { b } \right| }^{ 2 }-2\vec { a } .\vec { b } \\ \Rightarrow 4\vec { a } .\vec { b } =0\\ \therefore \quad \vec { a } .\vec { b } =0\\ Hence,\quad \vec { a } \bot \vec { b } \)
If \(\vec a\) + \(\vec b\)+ \(\vec c\) are unit vectors such that \(\vec a\) + \(\vec b\)+ \(\vec c\) = 0, then the value of \(\vec a.\vec b+\vec b.\vec c+\vec c.\vec a\) is
- (a)
1
- (b)
3
- (c)
-3/2
- (d)
none of these
\((\vec a+\vec b+\vec b)^2\) = 0
∴ \((\vec a)^2+(\vec b)^2+(\vec c)^2+2(\vec a.\vec b+\vec b.\vec c+\vec c.\vec a)=0\)
⇒ \(\vec a.\vec b+\vec b.\vec c+\vec c.\vec a=-\frac{1}{2}(1+1+1)\)
=\(-\frac{3}{2}\)
If \(\vec { a } ,\quad \vec { b } \quad and\quad \vec { c } \) be any three non-coplanar vectors. Then system \(\vec { a' } ,\) \(\vec { b' } \) and \(\vec { c' } \) which satisfies \(\vec { a } .\vec { a' } =\vec { b } .\vec { b' } =\vec { c } .\vec { c' } =1\) and \(\vec { a } .\vec { b' } =\vec { a } .\vec { c' } =\vec { b } .\vec { a' } =\vec { b } .\vec { c' } =\vec { c } .\vec { a' } =\vec { c } .\vec { b' } =0\) is called the reciprocal system to the vectors \(\vec { a } ,\quad \vec { b } \quad and\quad \vec { c } \) . If value of \(\vec { a' } \times \vec { a' } +\vec { b' } \times \vec { b' } +\vec { b' } \times \vec { c' } \) is
- (a)
0
- (b)
1
- (c)
\(\frac { \left( \vec { a } +\vec { b } +\vec { c } \right) }{ \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] } \)
- (d)
\(\frac { \left( \vec { a' } +\vec { b' } +\vec { c' } \right) }{ \left[ \begin{matrix} \vec { a' } & \vec { b' } & \vec { c' } \end{matrix} \right] } \)
By definition of reciprocal system
\(\vec { b } .\vec { a' } =\vec { c } .\vec { a' } =0\)
\(ie,\quad \vec { a' } \bot \vec { b } \quad and\quad \vec { a' } \bot \vec { c } \)
\(\Rightarrow \quad \vec { a' } ||\left( \vec { b } \times \vec { c } \right) \)
\(\therefore \quad \vec { a' } =t(\vec { b } \times \vec { c } )\quad \quad \quad ...(i)\)
Also, \(\vec { a } .\vec { a' } =1\)
\(\Rightarrow \quad 1=\vec { a' } .t(\vec { b } \times \vec { c } )\)
\(\therefore \quad t=\frac { 1 }{ \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] } \)
Now, from Eq. (i), \(\vec { a' } =\frac { (\vec { b } \times \vec { c } ) }{ \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] } \)
Similarly, \(\vec { b' } =\frac { (\vec { c } \times \vec { a } ) }{ \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] } \quad and\quad \vec { c' } =\frac { (\vec { a } \times \vec { b } ) }{ \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] } \)
\(\therefore \quad \vec { a' } \times \vec { b' } =\frac { (\vec { b } \times \vec { c } )\times (\vec { c } \times \vec { a } ) }{ { \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] }^{ 2 } } \)
\(=\frac { (\vec { b } \times \vec { c } .\vec { a } )\vec { c } \times (\vec { b } \times \vec { c } .\vec { c } )\vec { a } }{ { \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] }^{ 2 } } \)
\(=\frac { \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] \vec { c } -0 }{ { \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] }^{ 2 } } \)
\(=\frac { \vec { c } }{ { \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] } } \)
Similarly, \(\vec { b' } \times \vec { c' } =\frac { \vec { a } }{ { \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] } } \)
and \(\vec { c' } \times \vec { a' } =\frac { \vec { b } }{ { \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] } } \)
If \(\vec { a } ,\quad \vec { b } \quad and\quad \vec { c } \) be any three non-coplanar vectors. Then system \(\vec { a' } ,\) \(\vec { b' } \) and \(\vec { c' } \) which satisfies \(\vec { a } .\vec { a' } =\vec { b } .\vec { b' } =\vec { c } .\vec { c' } =1\) and \(\vec { a } .\vec { b' } =\vec { a } .\vec { c' } =\vec { b } .\vec { a' } =\vec { b } .\vec { c' } =\vec { c } .\vec { a' } =\vec { c } .\vec { b' } =0\) is called the reciprocal system to the vectors \(\vec { a } ,\quad \vec { b } \quad and\quad \vec { c } \) . The value of \(\left[ \vec { a } \vec { b } \vec { c } \right] \left\{ (\vec { a' } .\vec { a' } )\vec { a } +(\vec { a' } .\vec { b' } )\vec { b } +(\vec { a' } .\vec { c' } )\vec { c } \right\} \)is
- (a)
0
- (b)
\(\vec { a } \times \vec { b } \)
- (c)
\(\vec { b } \times \vec { c } \)
- (d)
\({ \left[ \vec { a } \vec { b } \vec { c } \right] }^{ -2 }\)
By definition of reciprocal system
\(\vec { b } .\vec { a' } =\vec { c } .\vec { a' } =0\)
\(ie,\quad \vec { a' } \bot \vec { b } \quad and\quad \vec { a' } \bot \vec { c } \)
\(\Rightarrow \quad \vec { a' } ||\left( \vec { b } \times \vec { c } \right) \)
\(\therefore \quad \vec { a' } =t(\vec { b } \times \vec { c } )\quad \quad \quad ...(i)\)
Also, \(\vec { a } .\vec { a' } =1\)
\(\Rightarrow \quad 1=\vec { a' } .t(\vec { b } \times \vec { c } )\)
\(\therefore \quad t=\frac { 1 }{ \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] } \)
Now, from Eq. (i), \(\vec { a' } =\frac { (\vec { b } \times \vec { c } ) }{ \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] } \)
Similarly, \(\vec { b' } =\frac { (\vec { c } \times \vec { a } ) }{ \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] } \quad and\quad \vec { c' } =\frac { (\vec { a } \times \vec { b } ) }{ \left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] } \)
Let \(\vec { b } \times \vec { c } =x\vec { a } +y\vec { b } +z\vec { c } \quad \quad ...(i)\)
\(\therefore \quad \left( \vec { b } \times \vec { c } \right) .\left( \vec { b } \times \vec { c } \right) =x\left( \vec { b } \times \vec { c } \right) .\vec { a } +y\left( \vec { b } \times \vec { c } \right) .\vec { b } +z\left( \vec { b } \times \vec { c } \right) .\vec { c } \)
\({ \left[ \vec { a } \vec { b } \vec { c } \right] }^{ 2 }(\vec { a' } .\vec { a' } )=x\left[ \vec { a } \vec { b } \vec { c } \right] +0+0\)
\(\therefore \quad x=(\vec { a' } .\vec { a' } )\left[ \vec { a } \vec { b } \vec { c } \right] \)
Similarly, \(y=(\vec { a' } .\vec { b' } )\left[ \vec { a } \vec { b } \vec { c } \right] \)
and \(z=(\vec { a' } .\vec { c' } )\left[ \vec { a } \vec { b } \vec { c } \right] \)
From Eq. (i),
\(\vec { b } \times \vec { c } ={ \left[ \vec { a } \vec { b } \vec { c } \right] \left[ (\vec { a' } .\vec { a' } )\vec { a } +(\vec { a' } .\vec { b' } )\vec { b } +(\vec { a' } .\vec { c' } )\vec { c } \right] }\)
\(\overrightarrow{a}\) and \(\overrightarrow{b}\) are unit vectors and \(|\overrightarrow{b}|\) = 4 with \(\overrightarrow { a } \times \overrightarrow { b } =2\overrightarrow { a } \times \overrightarrow { c } \) The angle between \(\overrightarrow{a}\) and \(\overrightarrow{c}\) is cos-1 (1/4). Then \(\overrightarrow { b } -2\overrightarrow { c } =\lambda \overrightarrow { a } ,\) if \(\lambda\) is
- (a)
3
- (b)
1/4
- (c)
-4
- (d)
-1/4
Given, \(|\overrightarrow{a}|=1,|\overrightarrow{c}|=1\)
and \(|\overrightarrow{b}|=4\)
and \(\overrightarrow{a} \times \overrightarrow{b} = 2 \overrightarrow{a} \times \overrightarrow{c}\)
Since, angle between \(\overrightarrow{a}\) and \(\overrightarrow{c}\) is cos-1 (1/4)
\(\therefore\) \(\overrightarrow { a } .\overrightarrow { c } =|\overrightarrow { a } ||\overrightarrow { c } |\frac { 1 }{ 4 } =1.1.\frac { 1 }{ 4 } \)
\(\Rightarrow\) \(\overrightarrow{a}.\overrightarrow{c}=\frac{1}{4}\)
and \(\overrightarrow{b}-2\overrightarrow{c}=\lambda\overrightarrow{a}\)
or \(\overrightarrow{b}=2\overrightarrow{c}+\lambda\overrightarrow{a}\)
Squaring b2 = 4 c2 + \(\lambda^{2}\) a2 +4 \(\lambda\) \(\overrightarrow{c}.\overrightarrow{a}\)
\(\Rightarrow\) 16 = 4 + \(\lambda^{2}\) + \(\lambda\)
\(\Rightarrow\) \({\lambda}^{2}+\lambda+12 = 0\)
\(\therefore\) \(\left(\lambda + 4 \right)\left(\lambda - 3 \right)=0\)
\(\Rightarrow\) \(\lambda= -4,3\)
Let A be the given point whose position vector relative to an origin O be \(\vec { a } \) and \(\vec { ON } =\vec { n } \) Let \(\vec { r } \) be the position vector of any point P which lies on the plane and passing through A and perpendicular to ON. Then for any point P on the plane.
\(\vec { AP } .\vec { n } =0\)
\(\Rightarrow\) \(\left( \vec { r } -\vec { a } \right) .\vec { n } =0\)
\(\Rightarrow\) \(\vec { r } .\vec { n } =\vec { a } .\vec { n } \)
\(\Rightarrow\) \(\vec { r } .\hat { n } =p\)
Where P is perpendicular distance of the plane from origin .
The equation of the plane through the point \(\hat { i } +2\hat { j } -\hat { k } \) and perpendicular to the line of intersection of the planes \(\vec { r } .\left( 3\hat { i } -\hat { j } +\hat { k } \right) =1\) and \(\vec { r } .\left( \hat { i } +4\hat { j } -2\hat { k } \right) =2\) is
- (a)
\(\vec { r } .\left( 2\hat { i } +7\hat { j } -13\hat { k } \right) =29\)
- (b)
\(\vec { r } .\left( 2\hat { i } -7\hat { j } -13\hat { k } \right) =1\)
- (c)
\(\vec { r } .\left( 2\hat { i } -7\hat { j } +13\hat { k } \right) +25=0\)
- (d)
\(\vec { r } .\left( 2\hat { i } +7\hat { j } +13\hat { k } \right) =3\)
The line of intersection of the planes \(\vec { r } .\left( 3\hat { i } -\hat { j } +\hat { k } \right) =1\) and \(\vec { r } .\left( \hat { i } +4\hat { j } -2\hat { k } \right) =2\) is common to both the planes.
Therefore, it is perpendicular to mormals to the two planes ie, \(\vec { { n }_{ 1 } } =3\hat { i } -\hat { j } +\hat { k } \) and \(\vec { { n }_{ 2 } } =\hat { i } +4\hat { j } -2\hat { k } \) Hence, it is parallel to the vector \(\vec { { n }_{ 1 } } \times \vec { { n }_{ 2 } } =\left( 3\hat { i } -\hat { j } +\hat { k } \right) \times \left( \hat { i } +4\hat { j } -2\hat { k } \right) \)
\(=-2\hat { i } +7\hat { j } +13\hat { k } \)
\(\therefore\) Equation of the plane passing through \(\vec { a } =\hat { i } +2\hat { j } -\hat { k } \) and normal to the vector \(\vec { { n } } =\vec { { n }_{ 1 } } \times \vec { { n }_{ 2 } } =-2\hat { i } +7\hat { j } +13\hat { k } \) is \(\left( \vec { r } -\vec { a } \right) .\vec { { n } } =0\Rightarrow \vec { { r } } .\vec { { n } } -\vec { { a } } .\vec { { n } } =0\)
\(\Rightarrow \quad \quad \vec { { r } } \left( -2\hat { i } +7\hat { j } +13\hat { k } \right) +1=0\)
or \(\vec { { r } } \left( 2\hat { i } -7\hat { j } -13\hat { k } \right) =1\)
If the vectors \(\overrightarrow{b}\) = ( tan \(\alpha\), -1, \(2\sqrt{sin \alpha /2}\) ) and \(\overrightarrow{c}\) = \(\left( tan\quad \alpha ,tan\quad \alpha ,-\frac { 3 }{ \sqrt { sin\quad \alpha /2 } } \right) \) are orthogonal and a vector \(\overrightarrow{a}\) = ( 1, 3, sin 2\(\alpha\) ) makes an obtuse angle with the z-axis, then the value of \(\alpha\) is
- (a)
\(\alpha =\left( 4n+1 \right) \pi -{ tan }^{ -1 }2\)
- (b)
\(\alpha =\left( 4n+2 \right) \pi -{ tan }^{ -1 }2\)
- (c)
\(\alpha =\left( 4n+1 \right) \pi { +tan }^{ -1 }2\)
- (d)
\(\alpha =\left( 4n+2 \right) \pi +{ tan }^{ -1 }2\)
\(\because\) \(\overrightarrow{a}\) = (1, 3, sin 2\(\alpha\) )
makes an obtuse angle with the z-axis, therefore its z-component in negative.
\(\therefore\) sin 2\(\alpha\) < 0
\(\Rightarrow\) -1 \(\le\) sin 2\(\alpha\) < 0
But \(\overrightarrow{b}.\overrightarrow{c}=0\)
tan2 \(\alpha\) - tan \(\alpha\) - 6 = 0
\(\therefore\) ( tan \(\alpha\) - 3) ( tan \(\alpha\) + 2 ) = 0
\(\Rightarrow\) tan \(\alpha\) = 3, -2
Now, tan \(\alpha\) = 3
\(\Rightarrow\) sin 2\(\alpha\) = \(\frac{2 tan \alpha}{1+{tab}^{2}\alpha}=\frac{6}{1+9}=\frac{3}{5}>0\)
Impossible \(\because\) sin 2\(\alpha\) < 0
Now, if tan \(\alpha\) = -2
\(\Rightarrow\) sin 2\(\alpha\) \(=\frac{2 tan \alpha}{1+ {tan}^{2}\alpha}=\frac{-4}{1+4}=\frac{4}{5}<0\)
which is true.
\(\therefore\) tan 2\(\alpha\) > 0
\(\therefore\) 2\(\alpha\) is in third quardrant also \(\sqrt{sin \alpha /2}\)
is meaningful, if 0 < sin \(\alpha / 2 <1,\) then
\(\alpha =\left( 4n+1 \right) \pi -{ tan }^{ -1 }2\)
\(\alpha =\left( 4n+2 \right) \pi -{ tan }^{ -1 }2\)
If \(\vec { a } ,\vec { b } \) and \(\vec { c } \) are any three vectors, then \(\vec { a } \times \left( \vec { b } \times \vec { c } \right) =\left( \vec { a } \times \vec { b } \right) \times \vec { c } \) if and only if
- (a)
\(\vec { b } \) and \(\vec { c } \) are collinear
- (b)
\(\vec { a } \) and \(\vec { c } \) are collinear
- (c)
\(\vec { a } \) and \(\vec { b } \) are collinear
- (d)
none of these
\(\vec { a } \times \left( \vec { b } \times \vec { c } \right) =\left( \vec { a } \times \vec { b } \right) \times \vec { c } \)
\(\Rightarrow\) \(\left( \vec { a } .\vec { c } \right) \vec { b } -\left( \vec { a } .\vec { b } \right) \vec { c } =\left( \vec { a } .\vec { c } \right) \vec { b } -\left( \vec { b } .\vec { c } \right) \vec { a } \)
\(\Rightarrow\) \(\left( \vec { b } .\vec { c } \right) \vec { a } =\left( \vec { a } .\vec { b } \right) \vec { c } \)
\(\Rightarrow\) \(\vec { a } =\frac { \left( \vec { a } .\vec { b } \right) }{ \left( \vec { b } .\vec { c } \right) } \vec { c } \)
\(\therefore\) \(\vec { a } \) and \(\vec { c } \) are collinear.
The vector \(\hat{i}+x\hat{j}+3\hat{k}\) is rotated through an angle \(\theta\) and doubled in magnitude, then it becomes \(4\hat{i}+(4x-2)\hat{j}+2\hat{k}\) The values of x are
- (a)
\(- \frac {2}{3}\)
- (b)
\(\frac{1}{3}\)
- (c)
\(\frac{2}{3}\)
- (d)
2
Let \(\overrightarrow{\alpha}=\hat{i}+x \hat{j}+3\hat{k},\overrightarrow{\beta}=4\hat{i}+(4x-2)\hat{j}+2\hat{k}\)
Given, \(2 |\overrightarrow{\alpha}|=|\overrightarrow{\beta}|\)
\(\Rightarrow\) \(2\sqrt{10+{x}^{2}}= \sqrt{20+44(2x-1)^{2}}\)
\(\Rightarrow\) 10 + x2 = 5 + ( 4x2 - 4x + 1 )
\(\Rightarrow\) 3x2 - 4x - 4 = 0 \(\Rightarrow\) x = 2, \(-\frac{2}{3}\)
The equation of the plane containing the line \(\vec r=\vec a+k\vec b\) and perpendicular to the plane \(\vec r.\vec n=q\) is
- (a)
\((\vec r-\vec b).(\vec n\times \vec a)=0\)
- (b)
\((\vec r-\vec a).\{\vec n\times(\vec a\times\vec b)\}=0\)
- (c)
\((\vec r-\vec a).(\vec n\times \vec b)=0\)
- (d)
\((\vec r-\vec b).\{\vec n\times (\vec a\times \vec b)\}=0\)
Equation of plane containing the line
\(\vec r=\vec a+k\vec b\)
and perpendicular to the plane
\(\vec r.\vec n=q\)
\(\therefore \vec b\) and \(\vec n\) are parallel
\(\Rightarrow \bot \)vector to \(\vec n\ and \ \vec b\ is\ (\vec n\times\vec b)\)
Hence, required plane is
\((\vec r-\vec a).(\vec n\times \vec b)=0\)
The volume of the tetrahedron whose vertices are the points with position vectors \(\hat { i } -6\hat { j } +10\hat { k } ,-\hat { i } -3\hat { j } 7\hat { k } ,5\hat { i } -\hat { j } +\lambda \hat { k } \) and \(7\hat { i } -4\hat { j } +7\hat { k } \) cubic units then the value of \(\lambda\)is
- (a)
-1
- (b)
1
- (c)
-7
- (d)
7
Let A, B, C and D be the points with the given position vectors, Then \(\overrightarrow { AB } =-2\hat { i } +3\hat { j } -3\hat { k } ,\) \(\overrightarrow { AC } =4\hat { i } +5\hat { j } +\left( \lambda -10 \right) \hat { k } \) and \(\overrightarrow { AD } =3\hat { i } +2\hat { j } -3\hat { k } \) so that the volume of the tetrahedron is
\(\frac { 1 }{ 6 } |\left( \overrightarrow { AB } \times \overrightarrow { AC } \right) .\overrightarrow { AC } |=\frac { 1 }{ 6 } |\begin{vmatrix} -2 & 3 & -3 \\ 4 & 5 & \lambda -10 \\ 6 & 2 & -3 \end{vmatrix}|\)
\(=\frac { 1 }{ 6 } |\left( -2 \right) \left[ -15-2\left( \lambda -10 \right) \right] -3\left[ -12-6\left( \lambda -10 \right) \right] |-3\left( 8-30 \right) |\)
\(=\frac { 1 }{ 6 } |-88+22\lambda |=11\) (given)
This gives \(\lambda = 1\) or 7.
Find the magnitude of the vector \(3\hat { i } +2\hat { j } +12\hat { k } \)
- (a)
\(\sqrt { 157 } \)
- (b)
\(4\sqrt { 11 } \)
- (c)
\(\sqrt { 213 } \)
- (d)
\(9\sqrt { 3 } \)
Magnitude of \(3\hat { i } +2\hat { j } +12\hat { k } \)
=\(\sqrt { { 3 }^{ 2 }+{ 2 }^{ 2 }+\left( 12 \right) ^{ 2 } } =\sqrt { 157 } \)
If \(\vec { a } =\hat { i } +\hat { j } +\hat { k } ,\vec { b } =4\hat { i } +3\hat { j } +4\hat { k } \) and \(\vec { c } =5\hat { i } +\alpha\hat { j } -\beta\hat { k } \)are linearly dependent vectors of
- (a)
\(\alpha=1, \beta=-1\)
- (b)
\(\alpha=1, \beta=\pm1\)
- (c)
\(\alpha=-1, \beta=\pm1\)
- (d)
\(\alpha=\pm1, \beta=1\)
If \(\vec { a } ,\vec { b } ,\vec { c } \) are linearly dependent vectors, then \(\vec { c } \) should be linearation combination of \(\vec { a } \quad and\quad\vec { b } \).
Let \(\vec { c } =p\vec { a } +q\vec { b } \)
i.e.,\(\hat { i } +\alpha \hat { j } +\beta \hat { k } =p\left( \hat { i } +\hat { j } +\hat { k } \right) +q\left( 4\hat { i } +3\hat { j } +4\hat { k } \right) \)
Equating co efficients of \(\hat { i } ,\hat { j } ,\hat { k } \) we get
1=p+4q,\(\alpha\)=p+3q, \(\beta\)=p+4q
From first and third, \(\beta=1\)
Now, \(\left| \vec { c } \right| =\sqrt { 3 } \Rightarrow 1+\alpha ^{ 2 }+{ \beta }^{ 2 }=3\)
\(\Rightarrow 1+{ \alpha }^{ 2 }+1=3\Rightarrow \alpha =\pm 1\)
Hence, \(\alpha =\pm 1,\beta =1\)
The fiigure formed by the four points \(\hat { i } ,\hat { j } -\hat { k } \) , \(\hat { 2i } +3\hat { j } ,5\hat { j } -2\hat { k } amd\quad \hat { k } -\hat { j } \)
- (a)
Square
- (b)
rhombus
- (c)
rectangle
- (d)
None of these
Let \(\vec { P } =\hat { i } +\hat { j } -\hat { k } ,\vec { Q } =2\hat { i } +3\hat { j } ,\vec { R } =5\hat { j } -2\hat { k } \) and \(\vec { S } =-\hat { j } +\hat { k } \)
\(\therefore \quad \vec { PQ } =\hat { i } +2\hat { j } +\hat { k } \Rightarrow \left| \vec { PQ } \right| =\sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 6 } \)
\(\vec { QR } =-2\hat { i } +2\hat { j } -2\hat { k } \Rightarrow \left| \vec { QR } \right| =\sqrt { \left( -2 \right) ^{ 2 }+{ 2 }^{ 2 }+\left( -1 \right) ^{ 2 } } =2\sqrt { 3 } \)
and \(\vec { RS } =-2\hat { 6i } -3\hat { k } \Rightarrow \left| \vec { RS } \right| =\sqrt { 36+9 } =\sqrt { 45 } =3\sqrt { 5 } \)
Similarly \(\vec { SP } =\hat { i } +2\hat { j } -2\hat { k } \)
Which does not satisfy the conditions of square, rectangle and parallelogram
If \(\vec { a } \) and are unit vectors enclosing an angle \(\theta \) and \(\left| \vec { a } +\vec { b } \right| <1\), then
- (a)
\(\theta =\cfrac { \pi }{ 2 } \)
- (b)
\(\theta =\cfrac { \pi }{ 3 } \)
- (c)
\(\pi \ge \theta >\cfrac { 2\pi }{ 3 } \)
- (d)
\(\cfrac { \pi }{ 3 } <\theta <\cfrac { 2\pi }{ 3 } \)
\(\left| \vec { a } +\vec { b } \right| <1\Rightarrow \left| \vec { a } +\vec { b } \right| ^{ 2 }<1\)
\(\Rightarrow \quad \left| \vec { a } \right| ^{ 2 }+\left| \vec { b } \right| ^{ 2 }+2\vec { a } .\vec { b } <1\Rightarrow 1+1+2\vec { a } .\vec { b } <1\)
\(\Rightarrow \quad \vec { a } .\vec { b } <\left| \vec { a } \right| \left| \vec { b } \right| \cos { \theta <-\cfrac { 1 }{ 2 } } \)
\(\Rightarrow \quad -1\le \cos { \theta } <-\cfrac { 1 }{ 2 } \Rightarrow \pi \ge \theta >\cfrac { 2\pi }{ 3 } \)
If \(\vec { a } =\hat { i } +\hat { j } +\hat { k } \) , \(\vec { b } =\hat { i } +\hat { j } \), \(\vec { c } =\hat { i } \) and \(\quad \left( \vec { a } \times \vec { b } \right) \times \vec { c } =\lambda \vec { a } +\mu \vec { b } \) then \(\lambda +\mu \) is equal to
- (a)
0
- (b)
1
- (c)
2
- (d)
3
\(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{matrix} \right| =-\hat { i } +\hat { j } \)
\(\left( \vec { a } \times \vec { b } \right) \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ -1 & 1 & 0 \\ 1 & 0 & 0 \end{matrix} \right| \)
⇒ \(\lambda \vec { a } +\mu \vec { b } =\hat { i } \left( 0-0 \right) -\hat { j } \left( 0-0 \right) +\hat { k } \left( 0-1 \right) \)
⇒ \(\left( \lambda +\mu \right) \hat { i } +\left( \lambda +\mu \right) \hat { j } +\left( \lambda \right) \hat { k } =-\hat { k } \)
On comparing, we get λ = -1 and λ + μ = 0 ⇒μ = 1
the angle between the vectors \(\vec { a } =\hat { i } +\hat { j } +4\hat { k } \) and \(\vec { b } =\hat { i } -\hat { j } +4\hat { k } \) is
- (a)
900
- (b)
450
- (c)
300
- (d)
150