JEE Mathematics Vector Algebra MCQs Set 21

Answer: (c) \( \le|\vec a|^2|\vec b|^2 \)

Explanation:
1. \( \vec a\cdot\vec b=|\vec a||\vec b|\cos\theta \) and \( \cos^2\theta\le1 \).
2. So \( (\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2\cos^2\theta\le|\vec a|^2|\vec b|^2 \).

Teacher's Note:
1. This is the Cauchy-Schwarz inequality for vectors.
2. Equality holds only when the vectors are parallel.

Answer: (c) \( \frac{5\hat i+2\hat j+\hat k}{3} \)

Explanation:
1. \( \vec a\cdot\vec b=-1+4+1=4 \) and \( |\vec b|^2=6 \).
2. The component along b is \( \frac46\vec b=\frac23(-\hat i+2\hat j+\hat k) \).
3. Perpendicular component \( =\vec a-\frac23\vec b=\left(\frac53,\frac23,\frac13\right)=\frac{5\hat i+2\hat j+\hat k}{3} \).

Teacher's Note:
1. Subtract the vector projection from the vector.
2. Check: the result dotted with b should be zero: −5 + 4 + 1 = 0.

Answer: (a) \( \frac{-\hat i+2\hat j-\hat k}{6} \)

Explanation:
1. \( \vec a\cdot\vec b=2-6+3=-1 \) and \( |\vec b|^2=6 \).
2. Orthogonal (vector) projection \( =\frac{\vec a\cdot\vec b}{|\vec b|^2}\vec b=-\frac16(\hat i-2\hat j+\hat k) \).

Teacher's Note:
1. The vector projection includes the direction of b, so it carries |b|² in the denominator.
2. Dividing by |b| only would give a scalar projection.

Answer: (d) \( \frac73 \)

Explanation:
1. \( \text{proj of }\vec a\text{ on }\vec b=\frac{\vec a\cdot\vec b}{|\vec b|} \) and \( \text{proj of }\vec b\text{ on }\vec a=\frac{\vec a\cdot\vec b}{|\vec a|} \).
2. So \( \lambda=\frac{|\vec a|}{|\vec b|} \). Here \( |\vec a|=7 \) and \( |\vec b|=3 \), so \( \lambda=\frac73 \).

Teacher's Note:
1. The common numerator a·b cancels (it is not zero here: −4 − 6 − 6 = −16).
2. The ratio of the projections is the ratio of the magnitudes.

Answer: (d) \( \frac32 \)

Explanation:
1. Let OP = (x, y, 0). The y-component is \( \sqrt3\cos60^\circ=\frac{\sqrt3}{2} \).
2. Then \( x^2=3-\frac34=\frac94 \), so \( |x|=\frac32 \).

Teacher's Note:
1. The component along an axis is magnitude × cosine of the angle with that axis.
2. Use Pythagoras in the plane for the other component.

Answer: (b) \( \frac{18(3\hat j+4\hat k)}{25} \)

Explanation:
1. \( \vec a\cdot\vec b=18 \) and \( |\vec b|^2=25 \).
2. Component \( =\frac{18}{25}(3\hat j+4\hat k) \).

Teacher's Note:
1. Vector component of a along b is (a·b/|b|²) b.
2. Compute a·b only from matching components (j with j): 6 × 3 = 18.

Answer: (b) 19

Explanation:
1. The dot product with \( \hat j \) picks the y-component.
2. y-component \( =2(1)+3(3)-4(-2)=2+9+8=19 \).

Teacher's Note:
1. Dot product with j gives the coefficient of j.
2. Be careful with the sign of −4 × (−2).

Answer: (b) \( \sqrt3 \)

Explanation:
1. Equal acute angles with the axes means the vector is \( \hat i+\hat j+\hat k \) (magnitude √3).
2. Projection \( =\frac{\vec a\cdot(\hat i+\hat j+\hat k)}{\sqrt3}=\frac{4-3+2}{\sqrt3}=\sqrt3 \).

Teacher's Note:
1. Equal direction cosines give a vector along (1, 1, 1).
2. The scalar projection divides by the magnitude of the vector projected on.

Answer: (c) \( \hat i+2\hat j+2\hat k \)

Explanation:
1. Let \( \vec a=\lambda\vec b \). Then \( \vec a\cdot\vec b=\lambda|\vec b|^2=81\lambda=27 \), so \( \lambda=\frac13 \).
2. So \( \vec a=\hat i+2\hat j+2\hat k \).

Teacher's Note:
1. Collinear means a is a scalar multiple of b.
2. |b|² = 9 + 36 + 36 = 81.

Answer: (d) \( 7\hat i+5\hat j+\hat k \)

Explanation:
1. x is parallel to the cross product \( (2,-3,1)\times(1,-2,3)=(-7,-5,-1) \), so \( \vec x=t(7,5,1) \).
2. Then \( \vec x\cdot(1,2,-7)=t(7+10-7)=10t=10 \), so t = 1.
3. Hence \( \vec x=7\hat i+5\hat j+\hat k \).

Teacher's Note:
1. A vector perpendicular to two vectors is parallel to their cross product.
2. Use the last condition to fix the scalar.

Answer: (b) equilateral

Explanation:
1. \( \overrightarrow{AB}=(1,1,-2),\ \overrightarrow{BC}=(1,-2,1),\ \overrightarrow{CA}=(-2,1,1) \).
2. Each has magnitude \( \sqrt6 \), so all three sides are equal.

Teacher's Note:
1. Compare the lengths of the three sides.
2. Also AB·BC = 1 − 2 − 2 = −3, which is not 0, so there is no right angle.

Answer: (d) \( \frac\pi3 \)

Explanation:
1. Perpendicular: \( \sin\theta-\sqrt3\cos\theta=0 \).
2. \( \tan\theta=\sqrt3 \), so \( \theta=\frac\pi3 \) for an acute angle.

Teacher's Note:
1. Set the dot product to zero.
2. tan θ = √3 gives 60° in the first quadrant.

Answer: (b) \( 5\sqrt2 \)

Explanation:
1. Adding the three conditions gives \( \vec A\cdot\vec B+\vec B\cdot\vec C+\vec C\cdot\vec A=0 \), and each condition individually gives all three pair products zero (e.g. A·B + A·C = 0, B·C + B·A = 0, C·A + C·B = 0 force each to 0).
2. So the vectors are mutually perpendicular and \( |\vec A+\vec B+\vec C|^2=9+16+25=50 \).
3. The magnitude is \( 5\sqrt2 \).

Teacher's Note:
1. Solve the three equations for the three dot products; the only solution is zero for all.
2. With zero cross terms, the squares simply add.

Answer: (b) 5

Explanation:
1. Parallelogram identity: \( |\vec a+\vec b|^2+|\vec a-\vec b|^2=2(|\vec a|^2+|\vec b|^2) \).
2. \( |\vec a+\vec b|^2=2(9+16)-25=25 \), so the magnitude is 5.

Teacher's Note:
1. The sum of the squares of the diagonals is twice the sum of the squares of the sides.
2. Here a ⊥ b since 9 + 16 = 25.

Answer: (b) 0

Explanation:
1. Expand each product: the first is \( ab-ac-db+dc \), the second \( bc-ba-dc+da \), the third \( ca-cb-da+db \) (writing xy for the dot product).
2. Every term cancels with its negative, so the sum is the scalar 0.

Teacher's Note:
1. The result of a dot product is a scalar, so the answer is the number 0, not the null vector.
2. Collect terms like a·b with b·a: they are equal.

Answer: (b) 9

Explanation:
1. The sum equals \( 6-2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a) \).
2. Since \( |\vec a+\vec b+\vec c|^2=3+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)\ge0 \), the bracket is at least \( -\frac32 \).
3. So the sum is at most \( 6+3=9 \).

Teacher's Note:
1. Use the non-negativity of |a + b + c|².
2. Equality holds when a + b + c = 0, e.g. three unit vectors at 120° to each other.

Answer: (a) \( \left(1,\frac12,-\frac12\right) \)

Explanation:
1. Let \( \vec b=t\vec a \). Then \( \vec a\cdot\vec b=t(4+1+1)=6t=3 \), so \( t=\frac12 \).
2. So \( \vec b=\left(1,\frac12,-\frac12\right) \).

Teacher's Note:
1. Collinear means b is a scalar multiple of a.
2. Check: a·b = 2 + 1/2 + 1/2 = 3.

Answer: (b) \( \sqrt3 \)

Explanation:
1. Let the unit vectors be a, b. \( |\vec a+\vec b|^2=2+2\vec a\cdot\vec b=1 \), so \( \vec a\cdot\vec b=-\frac12 \).
2. \( |\vec a-\vec b|^2=2-2\vec a\cdot\vec b=3 \), so the magnitude is \( \sqrt3 \).

Teacher's Note:
1. Use |a ± b|² = 2 ± 2a·b for unit vectors.
2. This is the parallelogram identity: the two squares add to 4.

Answer: (c) \( \frac\pi3 \)

Explanation:
1. The three vectors form an equilateral triangle: \( |\vec a-\vec b|^2=2|\vec a|^2-2\vec a\cdot\vec b=|\vec a|^2 \).
2. So \( \vec a\cdot\vec b=\frac{|\vec a|^2}{2} \) and \( \cos\theta=\frac12 \).

Teacher's Note:
1. Equal magnitudes of a, b and a − b make an equilateral triangle.
2. cos θ = 1/2 gives 60°.

Answer: (c) \( \frac12|\vec a-\vec b| \)

Explanation:
1. \( |\vec a-\vec b|^2=2-2\cos\theta=4\sin^2\frac\theta2 \).
2. So \( \sin\frac\theta2=\frac12|\vec a-\vec b| \) (the angle is between 0 and π, so the sine is non-negative).

Teacher's Note:
1. Use 1 − cos θ = 2 sin²(θ/2).
2. For |a + b| you would get cos(θ/2) instead.

Answer: (c) −1

Explanation:
1. \( |\vec a+\vec b+\vec c|^2=3+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=1 \).
2. With \( \vec a\cdot\vec b=0 \): \( \cos\beta+\cos\alpha=-1 \).

Teacher's Note:
1. For unit vectors, b·c = cos β and c·a = cos α.
2. Expand |a + b + c|² and use a ⊥ b.

Answer: (c) \( \cos^{-1}\frac1{\sqrt3} \)

Explanation:
1. Let the common magnitude be m. \( (\vec a+\vec b+\vec c)\cdot\vec a=m^2 \) and \( |\vec a+\vec b+\vec c|=m\sqrt3 \).
2. \( \cos\phi=\frac{m^2}{m\sqrt3\cdot m}=\frac1{\sqrt3} \).

Teacher's Note:
1. Cross terms vanish for perpendicular vectors.
2. This is the angle between a cube's diagonal and an edge.

Answer: (b) \( \cos^{-1}\sqrt{\frac{6}{41}} \)

Explanation:
1. Angle B is between \( \overrightarrow{BA}=(1,2,6) \) and \( \overrightarrow{BC}=(2,-1,1) \).
2. \( \cos B=\frac{2-2+6}{\sqrt{41}\sqrt6}=\frac{6}{\sqrt{246}}=\sqrt{\frac{6}{41}} \).

Teacher's Note:
1. The angle at B uses vectors pointing away from B: BA and BC, not AB and BC.
2. 6/√246 simplifies to √(36/246) = √(6/41).

Answer: (b) \( \frac\pi2 \)

Explanation:
1. \( \overrightarrow{BD}=\vec a-3\vec b \) and \( \overrightarrow{AC}=\vec a+3\vec b \).
2. \( \overrightarrow{BD}\cdot\overrightarrow{AC}=|\vec a|^2-9|\vec b|^2=0 \).
3. So they are perpendicular: the angle is π/2.

Teacher's Note:
1. Subtract position vectors to get BD and AC.
2. The difference-of-squares form vanishes since |a| = 3|b|.

Answer: (b) \( \frac2{\sqrt{17}} \)

Explanation:
1. For the plane \( Ax+By+Cz=d \), the distance from the origin is \( \frac{|d|}{\sqrt{A^2+B^2+C^2}} \).
2. \( \frac{2}{\sqrt{9+4+4}}=\frac2{\sqrt{17}} \).

Teacher's Note:
1. Normalise the normal vector: |n| = √17.
2. The distance equals |d|/|n|.