Download JEE MCQs for JEE Mathematics: Vector Algebra
Review structured MCQ sets for JEE Mathematics Vector Algebra. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.
Chapter-wise Objective Questions: Vector Algebra
View or download the dedicated Vector Algebra MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.
Multiple Choice Questions
(a) \( r^2\operatorname{cosec}\frac B2\operatorname{cosec}\frac C2\sin\frac A2 \)
(b) \( -r^2\operatorname{cosec}\frac A2\operatorname{cosec}\frac B2\sin\frac C2 \)
(c) \( r^2\operatorname{cosec}\frac B2\operatorname{cosec}\frac C2 \)
(d) \( -r^2\operatorname{cosec}\frac B2\operatorname{cosec}\frac C2\sin\frac A2 \)
Show Answer & Explanation
Answer: (d) \( -r^2\operatorname{cosec}\frac B2\operatorname{cosec}\frac C2\sin\frac A2 \)
Explanation:
1. \( IB=r\operatorname{cosec}\frac B2 \) and \( IC=r\operatorname{cosec}\frac C2 \).
2. \( \angle BIC=90^\circ+\frac A2 \), so \( \cos\angle BIC=-\sin\frac A2 \).
3. \( \vec{IB}\cdot\vec{IC}=-r^2\operatorname{cosec}\frac B2\operatorname{cosec}\frac C2\sin\frac A2 \).
Teacher's Note:
1. Use \( \angle BIC=90^\circ+\frac A2 \).
(a) \( \frac\pi3 \)
(b) \( \frac\pi4 \)
(c) \( \cos^{-1}\frac1{\sqrt3} \)
(d) \( \cos^{-1}\frac13 \)
Show Answer & Explanation
Answer: (d) \( \cos^{-1}\frac13 \)
Explanation:
1. In a face the centroid is at distance \( \frac{\sqrt3}{6} \) from each edge, and the height of a face is \( \frac{\sqrt3}{2} \).
2. The apex projects onto the centroid of the opposite face, so \( \cos\theta=\frac{\sqrt3/6}{\sqrt3/2}=\frac13 \).
Teacher's Note:
1. Compare the inradius of a face with the height of a face.
(a) \( 2\,\vec{SG} \)
(b) \( 3\,\vec{SG} \)
(c) \( \vec{SG} \)
(d) \( \frac12\vec{SG} \)
Show Answer & Explanation
Answer: (b) \( 3\,\vec{SG} \)
Explanation:
1. \( \vec{SA}+\vec{SB}+\vec{SC}=(\vec a+\vec b+\vec c)-3\vec s=3(\vec g-\vec s)=3\vec{SG} \).
Teacher's Note:
1. \( \vec g=\frac{\vec a+\vec b+\vec c}{3} \).
(a) \( 2\,\vec{AO} \)
(b) \( 2\,\vec{OO'} \)
(c) \( \frac12\vec{OO'} \)
(d) \( \vec{OO'} \)
Show Answer & Explanation
Answer: (b) \( 2\,\vec{OO'} \)
Explanation:
1. Take O as origin: \( \vec{OO'}=\vec a+\vec b+\vec c \) (orthocentre).
2. \( \vec{AO'}+\vec{BO'}+\vec{CO'}=3(\vec a+\vec b+\vec c)-(\vec a+\vec b+\vec c)=2(\vec a+\vec b+\vec c)=2\vec{OO'} \).
Teacher's Note:
1. With the circumcentre as origin the orthocentre is the sum of the vertex vectors.
Assertion (A): If a force \( \vec F \) passes through \( Q(\vec b) \), then the moment of the force about \( P(\vec a) \) is \( \vec F\times\vec r \) where \( \vec r=\vec{PQ} \).
Reason (R): Moment of a force is a vector.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. The moment about \( P \) is \( \vec r\times\vec F \) with \( \vec r=\vec{PQ} \), not \( \vec F\times\vec r \) (this has the opposite sign). So the assertion is false.
2. The moment (torque) is a vector, so the reason is true.
Teacher's Note:
1. The order in a cross product matters: moment \( =\vec r\times\vec F \).
Assertion (A): If \( \vec a=2\hat i+\hat k \), \( \vec b=3\hat j+4\hat k \) and \( \vec c=\lambda\vec a+\mu\vec b \) are coplanar, then \( \vec c=4\vec a-\vec b \).
Reason (R): A set of vectors \( \vec a_1,\dots,\vec a_n \) is linearly independent if every relation \( \ell_1\vec a_1+\dots+\ell_n\vec a_n=\vec 0 \) forces \( \ell_1=\dots=\ell_n=0 \).
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. \( \vec c=\lambda\vec a+\mu\vec b \) is coplanar with \( \vec a,\vec b \) for every \( \lambda,\mu \); coplanarity does not force \( \lambda=4,\mu=-1 \). The assertion is false.
2. The reason is the correct definition of linear independence, so it is true.
Teacher's Note:
1. Any linear combination of two vectors lies in their plane.
Assertion (A): If \( \cos\alpha,\cos\beta,\cos\gamma \) are the direction cosines of any line segment, then \( \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1 \).
Reason (R): If \( \cos\alpha,\cos\beta,\cos\gamma \) are the direction cosines of a line segment, then \( \cos2\alpha+\cos2\beta+\cos2\gamma=-1 \).
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. The assertion is the basic identity for direction cosines: true.
2. \( \cos2\alpha+\cos2\beta+\cos2\gamma=2(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)-3=-1 \): the reason is also true.
3. But the reason is a consequence of the assertion and does not explain why the sum of squares is \( 1 \).
Teacher's Note:
1. Check whether the reason actually explains the assertion, not just whether both are true.
(a) \( \hat i+\hat j \)
(b) \( \hat j \)
(c) \( \hat i \)
(d) \( \hat k \)
Show Answer & Explanation
Answer: (c) \( \hat i \)
Explanation:
1. A point on the line is \( (1+t,t,t) \). Then \( \vec r-(\hat i+\hat j)=(t,t-1,t) \).
2. Dot with \( (1,-1,-1) \): \( t-(t-1)-t=1-t=1 \), so \( t=0 \) and the point is \( \hat i \).
Teacher's Note:
1. Substitute the line into the plane equation.
(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: (d) outside \( \triangle OAB \) but inside \( \angle OBA \)
Explanation:
1. \( p=\frac34 \gt 0 \), \( q=-\frac12 \lt 0 \), \( p+q=\frac14 \lt 1 \): case (iv).
Teacher's Note:
1. Compute the signs of \( p \), \( q \) and \( p+q \) and match the case.
(a) \( -\hat i-2\hat j-2\hat k \)
(b) \( 3\hat i+2\hat j+2\hat k \)
(c) \( \hat i \)
(d) \( 2\hat i+\hat j+\hat k \)
Show Answer & Explanation
Answer: (d) \( 2\hat i+\hat j+\hat k \)
Explanation:
1. For the point with parameter \( t \): \( (\vec r-(\hat i+\hat j))\cdot n-1=1-t-1=-t \), so the distance is \( \frac{|t|}{\sqrt3} \).
2. \( \frac{|t|}{\sqrt3}=\frac1{\sqrt3} \) gives \( t=\pm1 \). \( t=1 \) gives \( 2\hat i+\hat j+\hat k \); \( t=-1 \) gives \( -\hat j-\hat k \), which is not offered.
Teacher's Note:
1. Distance from a plane is \( \frac{|\vec r\cdot\hat n-d|}{|n|} \).
(a) \( t^3+t^2+t=1 \) is a sufficient condition
(b) \( |\vec\alpha|=0 \) is a necessary condition
(c) \( |\vec\alpha|=|\vec\beta| \) is a necessary condition
(d) \( |\vec\alpha||\vec\beta|=0 \) is a necessary condition
Show Answer & Explanation
Answer: (a) \( t^3+t^2+t=1 \) is a sufficient condition
Explanation:
1. \( |\vec\alpha||\vec\beta|\tan\frac\theta2=|\vec\alpha||\vec\beta|\cos\theta \), so (for non-zero vectors) \( t=\frac{1-t^2}{1+t^2} \).
2. \( t+t^3=1-t^2 \), i.e. \( t^3+t^2+t=1 \). If this holds, the equality holds, so it is sufficient. The equality does not need a zero vector or equal magnitudes.
Teacher's Note:
1. Write \( \cos\theta \) in terms of \( t=\tan\frac\theta2 \).
(a) \( \frac\pi4 \)
(b) \( \frac\pi2 \)
(c) \( \frac\pi6 \)
(d) \( \frac{3\pi}{4} \)
Show Answer & Explanation
Answer: (a) \( \frac\pi4 \)
Explanation:
1. Normal of \( P_1 \): \( (0,2,3)\times(0,4,-3)=(-18,0,0) \). Normal of \( P_2 \): \( (0,1,-1)\times(3,3,0)=(3,-3,-3) \).
2. \( \vec A\parallel n_1\times n_2=(0,-54,54)\parallel(0,-1,1) \).
3. \( \cos\theta=\frac{|-1-2|}{\sqrt2\cdot3}=\frac1{\sqrt2} \), so the (acute) angle is \( \frac\pi4 \).
Teacher's Note:
1. The line of intersection of two planes is along the cross product of their normals.
(a) \( \vec a+\vec b-\sqrt2(\vec a\times\vec b) \)
(b) \( \vec a+\vec b+\sqrt2(\vec a\times\vec b) \)
(c) \( \vec a+\vec b-\sqrt3(\vec a\times\vec b) \)
(d) \( \vec a+\vec b+2\sqrt2(\vec a\times\vec b) \)
Show Answer & Explanation
Answer: (d) \( \vec a+\vec b+2\sqrt2(\vec a\times\vec b) \)
Explanation:
1. Take \( \vec a=(1,0,0) \), \( \vec b=\left(\frac12,\frac{\sqrt3}{2},0\right) \). Then \( \vec C \) lies above the centroid of \( OAB \): \( \vec{OC}=\left(\frac12,\frac{\sqrt3}{6},\sqrt{\frac23}\right) \).
2. \( 3\vec{OC}=\left(\frac32,\frac{\sqrt3}{2},\sqrt6\right) \) and \( \vec a+\vec b=\left(\frac32,\frac{\sqrt3}{2},0\right) \), \( \vec a\times\vec b=\left(0,0,\frac{\sqrt3}{2}\right) \).
3. So \( 3\vec{OC}=\vec a+\vec b+k(\vec a\times\vec b) \) with \( k\frac{\sqrt3}{2}=\sqrt6 \), i.e. \( k=2\sqrt2 \).
Teacher's Note:
1. Place the tetrahedron in coordinates and match components along \( \vec a+\vec b \) and \( \vec a\times\vec b \).
Assertion (A): If I is the incentre of \( \triangle ABC \), then \( |BC|\,\vec{IA}+|CA|\,\vec{IB}+|AB|\,\vec{IC}=\vec 0 \).
Reason (R): The position vector of the centroid of \( \triangle ABC \) is \( \frac{\vec{OA}+\vec{OB}+\vec{OC}}{3} \).
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. The incentre divides the sides in the ratio of the adjacent sides, so the side-weighted sum of \( \vec{IA},\vec{IB},\vec{IC} \) vanishes: true.
2. The centroid formula is also true, but it concerns a different point and does not explain the incentre relation.
Teacher's Note:
1. The incentre weights the vertices by the opposite side lengths.
(a) \( 2 \) and \( \frac43 \)
(b) \( -2 \) and \( \frac23 \)
(c) \( 2 \) and \( -\frac23 \)
(d) \( 1 \) and \( -\frac23 \)
Show Answer & Explanation
Answer: (c) \( 2 \) and \( -\frac23 \)
Explanation:
1. The new vector has twice the magnitude: \( 16+(4x-2)^2+4=4(1+x^2+9) \).
2. \( 16x^2-16x+24=4x^2+40 \), so \( 3x^2-4x-4=0 \).
3. \( (3x+2)(x-2)=0 \), giving \( x=2 \) or \( x=-\frac23 \).
Teacher's Note:
1. Equate the squared magnitudes.
(a) \( \frac1{3\sqrt2} \)
(b) \( \frac1{2\sqrt2} \)
(c) \( \frac1{6\sqrt2} \)
(d) \( \frac1{\sqrt2} \)
Show Answer & Explanation
Answer: (c) \( \frac1{6\sqrt2} \)
Explanation:
1. Take \( \vec{AB}=(1,0,0) \), \( \vec{AC}=\left(\frac12,\frac{\sqrt3}2,0\right) \), \( \vec{AD}=\left(\frac12,\frac{\sqrt3}6,\sqrt{\frac23}\right) \).
2. Volume \( =\frac16\left|[\vec{AB}\ \vec{AC}\ \vec{AD}]\right|=\frac16\cdot\frac{\sqrt3}{2}\sqrt{\frac23}=\frac16\cdot\frac1{\sqrt2} \).
Teacher's Note:
1. Use the scalar triple product for the volume.
Assertion (A): If three points P, Q, R have position vectors \( \vec a,\vec b,\vec c \) and \( 2\vec a+3\vec b-5\vec c=\vec 0 \), then P, Q, R are collinear.
Reason (R): For three points A, B, C, if \( \vec{AB}=\lambda\vec{AC} \) then A, B, C are collinear.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. \( \vec c=\frac{2\vec a+3\vec b}{5} \) with \( \frac25+\frac35=1 \), so R divides PQ in the ratio \( 3:2 \): P, Q, R are collinear.
2. \( \vec{AB}=\lambda\vec{AC} \) means the two vectors are parallel and share A, so the points are collinear. This is the principle used in the assertion.
Teacher's Note:
1. Coefficients summing to zero in a vector relation signals collinearity.
(i) \( \frac17(3\hat i+6\hat j-2\hat k) \)
(ii) \( \frac17(3\hat i-6\hat j-2\hat k) \)
(iii) \( \frac1{\sqrt{69}}(\hat i+2\hat j+8\hat k) \)
(iv) \( \frac1{\sqrt{69}}(-\hat i-2\hat j+8\hat k) \)
(a) (i) and (iv) only
(b) (iii) and (iv) only
(c) (ii) and (iii) only
(d) (i) and (ii) only
Show Answer & Explanation
Answer: (a) (i) and (iv) only
Explanation:
1. The diagonals are the sum \( (3,6,-2) \) and the difference \( (1,2,-8) \) of the sides.
2. \( |(3,6,-2)|=7 \) and \( |(1,2,-8)|=\sqrt{69} \), so the unit vectors are \( \pm\frac17(3,6,-2) \) and \( \pm\frac1{\sqrt{69}}(1,2,-8) \).
3. (i) is the first, (iv) is \( -\frac{1}{\sqrt{69}}(1,2,-8) \). (ii) and (iii) are not parallel to either diagonal.
Teacher's Note:
1. Diagonals of a parallelogram are the sum and the difference of its adjacent sides.
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Practice MCQs for JEE Mathematics Vector Algebra
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FAQs
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