JEE Mathematics Vector Algebra MCQs Set 26

Multiple Choice Questions (MCQs) for JEE Mathematics: Vector Algebra

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Practice Vector Algebra MCQs for JEE Mathematics

Access the complete set of multiple-choice questions for Vector Algebra below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: Let \( \vec x,\vec y,\vec z \) be vectors each of magnitude \( \sqrt2 \) with \( \frac\pi3 \) between every pair. \( \vec a \) is a non-zero vector perpendicular to \( \vec x \) and \( \vec y\times\vec z \), and \( \vec b \) is a non-zero vector perpendicular to \( \vec y \) and \( \vec z\times\vec x \). Which of the following are true?
(i) \( \vec b=(\vec b\cdot\vec z)(\vec z-\vec x) \)
(ii) \( \vec a=(\vec a\cdot\vec y)(\vec y-\vec z) \)
(iii) \( \vec a\cdot\vec b=-(\vec a\cdot\vec y)(\vec b\cdot\vec z) \)
(iv) \( \vec a=(\vec a\cdot\vec y)(\vec z-\vec y) \)
(a) (ii), (iii) and (iv)
(b) (i), (ii) and (iii)
(c) (i) and (iii) only
(d) (i), (ii) and (iv)
Show Answer & Explanation

Answer: (b) (i), (ii) and (iii)

Explanation:
1. All dot products between different vectors equal \( \sqrt2\cdot\sqrt2\cos60^\circ=1 \) and \( \vec x\cdot\vec x=2 \).
2. \( \vec a\parallel\vec x\times(\vec y\times\vec z)=\vec y(\vec x\cdot\vec z)-\vec z(\vec x\cdot\vec y)=\vec y-\vec z \). So \( \vec a=\lambda(\vec y-\vec z) \) and \( \vec a\cdot\vec y=\lambda(2-1)=\lambda \): (ii) true, (iv) false.
3. Similarly \( \vec b\parallel\vec y\times(\vec z\times\vec x)=\vec z-\vec x \) and \( \vec b\cdot\vec z=\mu \): (i) true.
4. \( \vec a\cdot\vec b=\lambda\mu(\vec y-\vec z)\cdot(\vec z-\vec x)=\lambda\mu(1-1-2+1)=-\lambda\mu \): (iii) true.

Teacher's Note:
1. Use the vector triple product expansion to find the direction of each vector.

Question 2: Let O be the origin, \( \vec{OA}=\vec a \), \( \vec{OB}=\vec b \) and \( \vec{OC}=p\vec a+q\vec b \). The point C lies (i) inside \( \triangle OAB \) if \( p \gt 0 \), \( q \gt 0 \), \( p+q \lt 1 \); (ii) outside \( \triangle OAB \) but inside \( \angle AOB \) if \( p \gt 0 \), \( q \gt 0 \), \( p+q \gt 1 \); (iii) outside \( \triangle OAB \) but inside \( \angle OAB \) if \( p \lt 0 \), \( q \gt 0 \), \( p+q \lt 1 \); (iv) outside \( \triangle OAB \) but inside \( \angle OBA \) if \( p \gt 0 \), \( q \lt 0 \), \( p+q \lt 1 \). For \( \frac12\vec{OC}=\frac12\vec a+\frac23\vec b \), the point C lies
(a) outside \( \triangle OAB \) but inside \( \angle AOB \)
(b) outside \( \triangle OAB \) but inside \( \angle OBA \)
(c) outside \( \triangle OAB \) but inside \( \angle OAB \)
(d) inside \( \triangle OAB \)
Show Answer & Explanation

Answer: (a) outside \( \triangle OAB \) but inside \( \angle AOB \)

Explanation:
1. \( \vec c=\vec a+\frac43\vec b \): \( p=1 \gt 0 \), \( q=\frac43 \gt 0 \), \( p+q=\frac73 \gt 1 \): case (ii).

Teacher's Note:
1. Compute the signs of \( p \), \( q \) and \( p+q \) and match the case.

Question 3: Let ABC be a triangle with \( BC=\alpha \), \( CA=\beta \), \( AB=\gamma \) (side lengths) and let \( \vec a,\vec b,\vec c \) be the position vectors of A, B, C. The position vector of the incentre I is
(a) \( \frac{\alpha\vec a+\beta\vec b+\gamma\vec c}{3} \)
(b) \( \frac{\vec a+\vec b+\vec c}{3} \)
(c) \( \frac{\alpha\vec a+\beta\vec b+\gamma\vec c}{\alpha+\beta+\gamma} \)
(d) \( \frac{\alpha\vec a+\beta\vec b+\gamma\vec c}{2} \)
Show Answer & Explanation

Answer: (c) \( \frac{\alpha\vec a+\beta\vec b+\gamma\vec c}{\alpha+\beta+\gamma} \)

Explanation:
1. The bisector AD divides BC in the ratio \( \frac{BD}{DC}=\frac{AB}{AC}=\frac\gamma\beta \), so \( \vec D=\frac{\beta\vec b+\gamma\vec c}{\beta+\gamma} \).
2. I divides AD in the ratio \( AI:ID=(\beta+\gamma):\alpha \), giving \( \vec I=\frac{\alpha\vec a+\beta\vec b+\gamma\vec c}{\alpha+\beta+\gamma} \).

Teacher's Note:
1. Each vertex is weighted by the length of the opposite side.

Question 4: If \( \vec a,\vec b,\vec c \) are non-coplanar vectors and \( \mu \) is a real number, the vectors \( \vec a+2\vec b+3\vec c \), \( \lambda\vec b+\mu\vec c \) and \( (2\lambda-1)\vec c \) are coplanar for
(a) all values of \( \lambda \) and \( \mu \)
(b) \( \lambda=0 \) only
(c) \( \lambda=0 \) or \( \lambda=\frac12 \), whatever \( \mu \) is
(d) no value of \( \lambda \)
Show Answer & Explanation

Answer: (c) \( \lambda=0 \) or \( \lambda=\frac12 \), whatever \( \mu \) is

Explanation:
1. Coplanar means the determinant of the coefficients in terms of \( \vec a,\vec b,\vec c \) is zero: \( \begin{vmatrix}1&2&3\\0&\lambda&\mu\\0&0&2\lambda-1\end{vmatrix}=\lambda(2\lambda-1) \).
2. \( \lambda(2\lambda-1)=0 \) gives \( \lambda=0 \) or \( \lambda=\frac12 \), for any \( \mu \).

Teacher's Note:
1. Three vectors are coplanar when the determinant of their coefficients in a non-coplanar basis is zero.

Question 5: Triangle ABC has position vectors \( 7\hat j+10\hat k \), \( -\hat i+6\hat j+6\hat k \) and \( -4\hat i+9\hat j+6\hat k \) for A, B, C. Which of the following are true?
(i) The triangle is isosceles
(ii) The triangle is equilateral
(iii) The triangle is right angled
(iv) The triangle is scalene
(a) (i) and (iii) only
(b) (ii) and (iii) only
(c) (i) and (iv) only
(d) (iii) and (iv) only
Show Answer & Explanation

Answer: (a) (i) and (iii) only

Explanation:
1. \( \vec{AB}=(-1,-1,-4) \), \( \vec{BC}=(-3,3,0) \), \( \vec{CA}=(4,-2,4) \).
2. \( |AB|^2=18 \), \( |BC|^2=18 \), \( |CA|^2=36=|AB|^2+|BC|^2 \).
3. Two sides are equal, so it is isosceles, and the Pythagoras relation holds, so it is right angled at B.

Teacher's Note:
1. Compute squared side lengths and test for equal sides and Pythagoras.

Question 6: For vectors \( \vec u,\vec v,\vec w \), which of the following expressions are meaningful?
(i) \( \vec u\cdot(\vec v\times\vec w) \)
(ii) \( (\vec u\times\vec v)\cdot\vec w \)
(iii) \( (\vec u\cdot\vec v)\vec w \)
(iv) \( \vec u\times(\vec v\cdot\vec w) \)
(a) (i), (ii) and (iii)
(b) (ii), (iii) and (iv)
(c) all four
(d) (i), (ii) and (iv)
Show Answer & Explanation

Answer: (a) (i), (ii) and (iii)

Explanation:
1. (i) and (ii) are scalar triple products.
2. (iii) is a scalar times a vector.
3. (iv) would need a cross product of a vector with the scalar \( \vec v\cdot\vec w \), which is not defined.

Teacher's Note:
1. A cross product needs two vectors; a dot product gives a scalar.

Question 7: If \( \vec a+2\vec b+3\vec c=\vec 0 \), then \( \vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a \) equals
(a) \( 6(\vec b\times\vec c) \)
(b) \( 2(\vec a\times\vec b) \)
(c) \( 3(\vec c\times\vec a) \)
(d) \( \vec 0 \)
Show Answer & Explanation

Answer: (a) \( 6(\vec b\times\vec c) \)

Explanation:
1. \( \vec a=-2\vec b-3\vec c \). Then \( \vec a\times\vec b=-3\vec c\times\vec b=3\,\vec b\times\vec c \) and \( \vec c\times\vec a=-2\,\vec c\times\vec b=2\,\vec b\times\vec c \).
2. Sum \( =3+1+2=6 \) times \( \vec b\times\vec c \).

Teacher's Note:
1. Eliminate one vector using the given relation.

Question 8: All values of \( \lambda \) for which \( x,y,z \) (not all zero) satisfy \( (\hat i+\hat j+3\hat k)x+(3\hat i-3\hat j+\hat k)y+(-4\hat i+5\hat j)z=\lambda(x\hat i+y\hat j+z\hat k) \) are
(a) \( \lambda=0 \) and \( \lambda=-1 \)
(b) \( \lambda=1 \) and \( \lambda=-1 \)
(c) \( \lambda=0 \) and \( \lambda=1 \)
(d) \( \lambda=0 \), \( \lambda=1 \) and \( \lambda=-1 \)
Show Answer & Explanation

Answer: (a) \( \lambda=0 \) and \( \lambda=-1 \)

Explanation:
1. Comparing components: \( x+3y-4z=\lambda x \), \( x-3y+5z=\lambda y \), \( 3x+y=\lambda z \).
2. This is \( (M-\lambda I)\begin{pmatrix}x\\y\\z\end{pmatrix}=0 \) with \( M=\begin{pmatrix}1&3&-4\\1&-3&5\\3&1&0\end{pmatrix} \).
3. A non-zero solution needs \( \det(M-\lambda I)=-\lambda(\lambda+1)^2=0 \), so \( \lambda=0 \) or \( \lambda=-1 \).

Teacher's Note:
1. A non-trivial solution exists only when the determinant of the coefficient matrix vanishes.

Question 9: The resolved part of \( \vec a \) along \( \vec b \) is \( \vec\lambda \) and that perpendicular to \( \vec b \) is \( \vec\mu \). Which of the following are correct?
(i) \( \vec\lambda=\frac{(\vec a\cdot\vec b)\vec a}{|\vec a|^2} \)
(ii) \( \vec\lambda=\frac{(\vec a\cdot\vec b)\vec b}{|\vec b|^2} \)
(iii) \( \vec\mu=\frac{(\vec b\cdot\vec b)\vec a-(\vec a\cdot\vec b)\vec b}{|\vec b|^2} \)
(iv) \( \vec\mu=\frac{\vec b\times(\vec a\times\vec b)}{|\vec b|^2} \)
(a) (ii), (iii) and (iv)
(b) (i) and (iv) only
(c) (ii) and (iii) only
(d) (i), (ii) and (iii)
Show Answer & Explanation

Answer: (a) (ii), (iii) and (iv)

Explanation:
1. The component of \( \vec a \) along \( \vec b \) is \( \frac{(\vec a\cdot\vec b)}{|\vec b|^2}\vec b \), so (ii) is right and (i) is wrong.
2. \( \vec\mu=\vec a-\vec\lambda=\frac{|\vec b|^2\vec a-(\vec a\cdot\vec b)\vec b}{|\vec b|^2} \), which is (iii).
3. \( \vec b\times(\vec a\times\vec b)=\vec a|\vec b|^2-\vec b(\vec a\cdot\vec b) \), so (iv) equals the same vector.

Teacher's Note:
1. Subtract the component along \( \vec b \) from \( \vec a \) to get the perpendicular part.

Question 10: A vector \( \vec a \) has components \( 2p \) and \( 1 \) with respect to a rectangular system. The system is rotated counter-clockwise about the origin and \( \vec a \) now has components \( p+1 \) and \( 1 \). The possible values of \( p \) are
(a) \( -1 \) and \( \frac13 \)
(b) \( 1 \) and \( 2 \)
(c) \( 1 \) and \( -\frac13 \)
(d) \( -1 \) and \( 2 \)
Show Answer & Explanation

Answer: (c) \( 1 \) and \( -\frac13 \)

Explanation:
1. Rotation of axes does not change the magnitude: \( 4p^2+1=(p+1)^2+1 \).
2. \( 3p^2-2p-1=0 \), so \( (3p+1)(p-1)=0 \) and \( p=1 \) or \( -\frac13 \).

Teacher's Note:
1. Magnitude of a vector is unchanged by rotating the coordinate axes.

Question 11: In triangle ABC with incentre I and inradius \( r \), the magnitude \( |\vec{IB}\times\vec{IC}| \) equals
(a) \( r^2\operatorname{cosec}\frac B2\operatorname{cosec}\frac C2 \)
(b) \( r^2\operatorname{cosec}\frac B2\operatorname{cosec}\frac C2\sin\frac A2 \)
(c) \( r^2\operatorname{cosec}\frac B2\operatorname{cosec}\frac C2\cos\frac A2 \)
(d) \( r^2\operatorname{cosec}\frac A2\operatorname{cosec}\frac B2\cos\frac C2 \)
Show Answer & Explanation

Answer: (c) \( r^2\operatorname{cosec}\frac B2\operatorname{cosec}\frac C2\cos\frac A2 \)

Explanation:
1. \( |\vec{IB}\times\vec{IC}|=IB\cdot IC\sin\angle BIC \) with \( \sin\left(90^\circ+\frac A2\right)=\cos\frac A2 \).
2. \( IB\cdot IC=r^2\operatorname{cosec}\frac B2\operatorname{cosec}\frac C2 \), giving the stated result.

Teacher's Note:
1. \( \sin(90^\circ+x)=\cos x \).

Question 12: A vector of magnitude \( 2 \) along the bisector of the angle between \( 2\hat i-2\hat j+\hat k \) and \( \hat i+2\hat j-2\hat k \) is
(a) \( \frac1{\sqrt{10}}(\hat i+3\hat k) \)
(b) \( \frac{2}{\sqrt{10}}(3\hat i-\hat k) \)
(c) \( \frac{1}{\sqrt{26}}(\hat i-4\hat j+3\hat k) \)
(d) \( \frac{2}{\sqrt{26}}(\hat i-4\hat j+3\hat k) \)
Show Answer & Explanation

Answer: (b) \( \frac{2}{\sqrt{10}}(3\hat i-\hat k) \)

Explanation:
1. Both vectors have magnitude \( 3 \). The bisector direction is the sum of the unit vectors: \( \frac13(3,0,-1) \).
2. Its unit vector is \( \frac{(3,0,-1)}{\sqrt{10}} \).
3. Magnitude \( 2 \) gives \( \frac2{\sqrt{10}}(3\hat i-\hat k) \).

Teacher's Note:
1. The bisector of two vectors is along the sum of their unit vectors.

Question 13: A new operation \( * \) is defined between two non-parallel vectors \( \vec\alpha,\vec\beta \) by \( \vec\alpha*\vec\beta=|\vec\alpha||\vec\beta|\tan\frac\theta2 \), where \( \theta \) is the angle between them. The condition for \( \vec\alpha \) and \( \vec\beta \) to be perpendicular is
(a) \( \frac{\vec\alpha*\vec\beta}{|\vec\alpha|}=1 \)
(b) \( \vec\alpha*\vec\beta=0 \)
(c) \( \frac{\vec\alpha*\vec\beta}{|\vec\alpha||\vec\beta|}=-1 \)
(d) \( \frac{\vec\alpha*\vec\beta}{|\vec\alpha||\vec\beta|}=1 \)
Show Answer & Explanation

Answer: (d) \( \frac{\vec\alpha*\vec\beta}{|\vec\alpha||\vec\beta|}=1 \)

Explanation:
1. Perpendicular means \( \theta=90^\circ \), so \( \tan\frac\theta2=\tan45^\circ=1 \).
2. Hence \( \frac{\vec\alpha*\vec\beta}{|\vec\alpha||\vec\beta|}=1 \).

Teacher's Note:
1. Divide the operation by the product of magnitudes to isolate \( \tan\frac\theta2 \).

Question 14: A new operation \( * \) is defined between two non-parallel vectors \( \vec\alpha,\vec\beta \) by \( \vec\alpha*\vec\beta=|\vec\alpha||\vec\beta|\tan\frac\theta2 \), where \( \theta \) is the angle between them. The value of \( \vec\alpha*\vec\alpha \) is
(a) \( |\vec\alpha|^2 \)
(b) not defined
(c) \( 0 \)
(d) \( |\vec\alpha|^2-3 \)
Show Answer & Explanation

Answer: (c) \( 0 \)

Explanation:
1. The angle of a vector with itself is \( 0 \), and \( \tan0=0 \), so \( \vec\alpha*\vec\alpha=0 \).

Teacher's Note:
1. The angle between a vector and itself is zero.

Question 15: Let O be the origin, \( \vec{OA}=\vec a \), \( \vec{OB}=\vec b \) and \( \vec{OC}=p\vec a+q\vec b \). The point C lies (i) inside \( \triangle OAB \) if \( p \gt 0 \), \( q \gt 0 \), \( p+q \lt 1 \); (ii) outside \( \triangle OAB \) but inside \( \angle AOB \) if \( p \gt 0 \), \( q \gt 0 \), \( p+q \gt 1 \); (iii) outside \( \triangle OAB \) but inside \( \angle OAB \) if \( p \lt 0 \), \( q \gt 0 \), \( p+q \lt 1 \); (iv) outside \( \triangle OAB \) but inside \( \angle OBA \) if \( p \gt 0 \), \( q \lt 0 \), \( p+q \lt 1 \). For \( \vec{OC}=\frac12\vec a+\frac13\vec b \), the point C lies
(a) inside \( \triangle OAB \)
(b) outside \( \triangle OAB \) but inside \( \angle OAB \)
(c) outside \( \triangle OAB \) but inside \( \angle AOB \)
(d) outside \( \triangle OAB \) but inside \( \angle OBA \)
Show Answer & Explanation

Answer: (a) inside \( \triangle OAB \)

Explanation:
1. \( p=\frac12 \gt 0 \), \( q=\frac13 \gt 0 \) and \( p+q=\frac56 \lt 1 \): case (i).

Teacher's Note:
1. Compute the signs of \( p \), \( q \) and \( p+q \) and match the case.

Question 16: Let O be the origin, \( \vec{OA}=\vec a \), \( \vec{OB}=\vec b \) and \( \vec{OC}=p\vec a+q\vec b \). The point C lies (i) inside \( \triangle OAB \) if \( p \gt 0 \), \( q \gt 0 \), \( p+q \lt 1 \); (ii) outside \( \triangle OAB \) but inside \( \angle AOB \) if \( p \gt 0 \), \( q \gt 0 \), \( p+q \gt 1 \); (iii) outside \( \triangle OAB \) but inside \( \angle OAB \) if \( p \lt 0 \), \( q \gt 0 \), \( p+q \lt 1 \); (iv) outside \( \triangle OAB \) but inside \( \angle OBA \) if \( p \gt 0 \), \( q \lt 0 \), \( p+q \lt 1 \). For \( \vec{OC}=-\frac12\vec a+\frac34\vec b \), the point C lies
(a) outside \( \triangle OAB \) but inside \( \angle OAB \)
(b) outside \( \triangle OAB \) but inside \( \angle AOB \)
(c) inside \( \triangle OAB \)
(d) outside \( \triangle OAB \) but inside \( \angle OBA \)
Show Answer & Explanation

Answer: (a) outside \( \triangle OAB \) but inside \( \angle OAB \)

Explanation:
1. \( p=-\frac12 \lt 0 \), \( q=\frac34 \gt 0 \), \( p+q=\frac14 \lt 1 \): case (iii).

Teacher's Note:
1. Compute the signs of \( p \), \( q \) and \( p+q \) and match the case.

Question 17: A vector \( \vec d \) is directed along the internal bisector of the angle between \( \vec b=7\hat i-4\hat j-4\hat k \) and \( \vec c=-2\hat i-\hat j+2\hat k \), with \( |\vec d|=5\sqrt6 \). Then \( \vec d \) is
(a) \( \frac53(\hat i+7\hat j+2\hat k) \)
(b) \( \frac53(\hat i-7\hat j+2\hat k) \)
(c) \( \frac53(5\hat i+5\hat j-2\hat k) \)
(d) \( \frac53(-5\hat i+5\hat j+2\hat k) \)
Show Answer & Explanation

Answer: (b) \( \frac53(\hat i-7\hat j+2\hat k) \)

Explanation:
1. \( |\vec b|=9 \), \( |\vec c|=3 \). \( \frac{\vec b}{9}+\frac{\vec c}{3}=\frac{(7,-4,-4)+(-6,-3,6)}{9}=\frac{(1,-7,2)}{9} \).
2. \( |(1,-7,2)|=\sqrt{54} \), so \( \vec d=5\sqrt6\cdot\frac{(1,-7,2)}{\sqrt{54}}=\frac53(1,-7,2) \).

Teacher's Note:
1. The internal bisector direction is \( \frac{\vec b}{|\vec b|}+\frac{\vec c}{|\vec c|} \).

Question 18: Let \( \vec a=2\hat i-\hat j+\hat k \), \( \vec b=\hat i+2\hat j-\hat k \) and \( \vec c=\hat i+\hat j-2\hat k \). A vector in the plane of \( \vec b \) and \( \vec c \) whose projection on \( \vec a \) has magnitude \( \sqrt{\frac23} \) is
(a) \( 2\hat i+\hat j+5\hat k \)
(b) \( -2\hat i-\hat j+5\hat k \)
(c) \( 2\hat i+3\hat j-3\hat k \)
(d) \( 2\hat i+3\hat j+3\hat k \)
Show Answer & Explanation

Answer: (b) \( -2\hat i-\hat j+5\hat k \)

Explanation:
1. A vector in the plane of \( \vec b \) and \( \vec c \) is perpendicular to \( \vec b\times\vec c=(-3,1,-1) \).
2. \( -2\hat i-\hat j+5\hat k \) gives \( 6-1-5=0 \): it lies in the plane. \( 2\hat i+3\hat j-3\hat k \) also lies in the plane but its projection on \( \vec a \) is \( -\frac{2\sqrt6}{3} \) in magnitude \( \frac{2\sqrt6}{3}\neq\sqrt{\frac23} \).
3. Projection of \( -2\hat i-\hat j+5\hat k \) on \( \vec a \): \( \frac{-4+1+5}{\sqrt6}=\frac{2}{\sqrt6}=\sqrt{\frac23} \).

Teacher's Note:
1. Check coplanarity with the triple product, then the projection \( \frac{\vec v\cdot\vec a}{|\vec a|} \).

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