JEE Mathematics Vector Algebra MCQs Set 20

Answer: (d) \( (\vec{a}·\vec{c})(\vec{b}·\vec{d}) \)

Explanation:
1. \( (\vec{a}×\vec{b})·(\vec{c}×\vec{d})=(\vec{a}·\vec{c})(\vec{b}·\vec{d})−(\vec{a}·\vec{d})(\vec{b}·\vec{c}) \).
2. Adding \( (\vec{b}·\vec{c})(\vec{a}·\vec{d}) \) cancels the second term.
3. The result is \( (\vec{a}·\vec{c})(\vec{b}·\vec{d}) \).

Teacher's Note:
1. Use the identity and see which term cancels.
2. Do not forget the order of the dot products.

Answer: (c) 0

Explanation:
1. \( \vec{a}×\vec{b} \) is normal to the plane of \( \vec{a},\vec{b} \) and \( \vec{c}×\vec{d} \) is normal to the plane of \( \vec{c},\vec{d} \).
2. Perpendicular planes have perpendicular normals.
3. So \( (\vec{a}×\vec{b})·(\vec{c}×\vec{d})=0 \).

Teacher's Note:
1. The angle between two planes equals the angle between their normals.
2. Perpendicular normals give a zero dot product.

Answer: (a) \( −70\hat{i}−28\hat{j}−64\hat{k} \)

Explanation:
1. \( \vec{a}×\vec{b}=(−10,9,7) \) and \( \vec{c}×\vec{d}=(1·2−1·(−4),\ 1·3−1·2,\ 1·(−4)−1·3)=(6,1,−7) \).
2. \( (−10,9,7)×(6,1,−7)=(−63−7,\ 42−70,\ −10−54)=(−70,−28,−64) \).
3. This is the first option.

Teacher's Note:
1. Compute the two cross products first and cross them.
2. Check signs in each component.

Answer: (d) \( −[\vec{a}\ \vec{b}\ \vec{c}] \)

Explanation:
1. Treat \( \vec{a}×\vec{b} \) as one vector \( \vec{u} \): \( \vec{u}×(\vec{c}×\vec{d})=(\vec{u}·\vec{d})\vec{c}−(\vec{u}·\vec{c})\vec{d} \).
2. So \( (\vec{a}×\vec{b})×(\vec{c}×\vec{d})=[\vec{a}\ \vec{b}\ \vec{d}]\vec{c}−[\vec{a}\ \vec{b}\ \vec{c}]\vec{d} \).
3. Comparing, \( m=−[\vec{a}\ \vec{b}\ \vec{c}] \).

Teacher's Note:
1. Use the triple product formula with \( \vec{a}×\vec{b} \) as the first vector.
2. The coefficient of \( \vec{d} \) carries a minus sign.

Answer: (a) \( \vec0 \)

Explanation:
1. A non-trivial relation means \( \vec{a},\vec{b},\vec{c} \) are coplanar, so \( [\vec{a}\ \vec{b}\ \vec{c}]=0 \).
2. \( (\vec{b}×\vec{c})×(\vec{c}×\vec{a})=[\vec{a}\ \vec{b}\ \vec{c}]\vec{c}=\vec0 \).
3. The whole expression is the null vector.

Teacher's Note:
1. Spot that the condition gives coplanarity.
2. Then the inner cross product has the factor \( [\vec{a}\ \vec{b}\ \vec{c}] \).

Answer: (d) \( 2(\vec{a}×\vec{b}) \)

Explanation:
1. Put \( \vec{u}=\vec{a}×\vec{b} \). Then \( \hat{i}×(\vec{u}×\hat{i})=\vec{u}−(\vec{u}·\hat{i})\hat{i} \).
2. Summing over the three axes: \( 3\vec{u}−\vec{u}=2\vec{u} \).
3. Hence the value is \( 2(\vec{a}×\vec{b}) \).

Teacher's Note:
1. Treat \( \vec{a}×\vec{b} \) as a single vector and use the standard summation.
2. The result is twice the vector.

Answer: (a) \( 90^\circ,60^\circ \)

Explanation:
1. \( (\vec{a}·\vec{c})\vec{b}−(\vec{a}·\vec{b})\vec{c}=\frac12\vec{b} \).
2. Since \( \vec{b},\vec{c} \) are non-collinear, \( \vec{a}·\vec{c}=\frac12 \) and \( \vec{a}·\vec{b}=0 \).
3. So \( (\vec{a},\vec{c})=60^\circ \) and \( (\vec{a},\vec{b})=90^\circ \).

Teacher's Note:
1. Compare the coefficients of the independent vectors \( \vec{b} \) and \( \vec{c} \).
2. Unit vectors give \( \cos \) directly from the dot product.

Answer: (a) \( \dfrac{3\pi}{4} \)

Explanation:
1. \( (\vec{a}·\vec{c})\vec{b}−(\vec{a}·\vec{b})\vec{c}=\frac{1}{\sqrt2}\vec{b}+\frac{1}{\sqrt2}\vec{c} \).
2. Since \( \vec{b},\vec{c} \) are independent, \( \vec{a}·\vec{c}=\frac1{\sqrt2} \) and \( \vec{a}·\vec{b}=−\frac1{\sqrt2} \).
3. Hence \( \cos θ=−\frac1{\sqrt2} \) and \( θ=\frac{3\pi}{4} \).

Teacher's Note:
1. Equate coefficients of \( \vec{b} \) and \( \vec{c} \) after expanding.
2. Non-coplanar vectors are linearly independent.

Answer: (b) \( 3/2 \)

Explanation:
1. \( \vec{a}×\vec{b}=(2,−2,1) \) has magnitude 3, and \( |\vec{a}|=3 \).
2. \( |\vec{c}−\vec{a}|^2=8 \) gives \( |\vec{c}|^2+9−2|\vec{c}|=8 \), so \( (|\vec{c}|−1)^2=0 \) and \( |\vec{c}|=1 \).
3. \( |(\vec{a}×\vec{b})×\vec{c}|=3·1·\sin30^\circ=\frac32 \).

Teacher's Note:
1. Use the given condition to find \( |\vec{c}| \) first.
2. The magnitude of a cross product is \( |\vec{u}||\vec{c}|\sin θ \).

Answer: (a) \( \vec{a}−\vec{b} \)

Explanation:
1. \( (\vec{a}+\vec{b})×(\vec{a}×\vec{b})=\{(\vec{a}+\vec{b})·\vec{b}\}\vec{a}−\{(\vec{a}+\vec{b})·\vec{a}\}\vec{b} \).
2. \( =(\vec{a}·\vec{b}+1)\vec{a}−(1+\vec{a}·\vec{b})\vec{b}=(1+\vec{a}·\vec{b})(\vec{a}−\vec{b}) \).
3. So it is parallel to \( \vec{a}−\vec{b} \).

Teacher's Note:
1. Use the triple product formula with \( \vec{u}=\vec{a}+\vec{b} \) as the first vector.
2. Unit vectors give \( |\vec{a}|^2=|\vec{b}|^2=1 \).

Answer: (a) \( α=−2, β=−4, γ=−\frac23 \)

Explanation:
1. \( \vec{a}·\vec{c}=2+1+2=5 \) and \( \vec{a}·\vec{b}=2+2+2=6 \).
2. \( \vec{a}×(\vec{b}×\vec{c})=5\vec{b}−6\vec{c}=(−1,4,−2) \).
3. So \( 1+α=−1 \) gives \( α=−2 \); \( β(−1)=4 \) gives \( β=−4 \); \( γ(−1)(−3)=−2 \) gives \( γ=−\frac23 \).

Teacher's Note:
1. Use the expansion \( \vec{a}×(\vec{b}×\vec{c})=(\vec{a}·\vec{c})\vec{b}−(\vec{a}·\vec{b})\vec{c} \) and substitute the components.
2. Then match components one by one.

Answer: (b) \( \vec{b}×\vec{a} \)

Explanation:
1. \( \vec{a}×(\vec{a}×\vec{b})=(\vec{a}·\vec{b})\vec{a}−\vec{b} \) for unit \( \vec{a} \).
2. \( \vec{a}×\{(\vec{a}·\vec{b})\vec{a}−\vec{b}\}=−\vec{a}×\vec{b} \).
3. That equals \( \vec{b}×\vec{a} \).

Teacher's Note:
1. Apply the triple product formula twice.
2. The first term vanishes because \( \vec{a}×\vec{a}=\vec0 \).

Answer: (a) \( \dfrac34 \)

Explanation:
1. \( (\vec{a}·\vec{c})\vec{a}−3\vec{c}+3\vec{b}=\vec0 \), so \( 3\vec{b}=3\vec{c}−(\vec{a}·\vec{c})\vec{a} \).
2. Take squared magnitudes, using \( |\vec{a}|^2=3 \): \( 9=9·4−6(\vec{a}·\vec{c})^2+3(\vec{a}·\vec{c})^2=36−3(\vec{a}·\vec{c})^2 \).
3. So \( (\vec{a}·\vec{c})^2=9 \). With \( \vec{a}·\vec{c}=2\sqrt3\cosθ \), \( 12\cos^2θ=9 \) and \( \cos^2θ=\frac34 \).

Teacher's Note:
1. Rearrange so that the vector \( \vec{b} \) is isolated and then take magnitudes.
2. Use \( \vec{a}·\vec{c}=|\vec{a}||\vec{c}|\cosθ \).

Answer: (a) 6

Explanation:
1. \( (\vec{a}×\hat{i})×\hat{j}=(\vec{a}·\hat{j})\hat{i}−(\hat{i}·\hat{j})\vec{a}=(\vec{a}·\hat{j})\hat{i} \), whose square is \( a_2^2 \).
2. Cyclically the sum is \( a_1^2+a_2^2+a_3^2=|\vec{a}|^2 \).
3. Here this is \( 1+1+4=6 \).

Teacher's Note:
1. Evaluate each bracket with the triple product formula; the second term is zero.
2. The cyclic sum of squared components is the squared magnitude.

Answer: (c) 3

Explanation:
1. \( |\vec{a}×\vec{b}|^2+(\vec{a}·\vec{b})^2=|\vec{a}|^2|\vec{b}|^2 \).
2. So \( 16|\vec{b}|^2=144 \) and \( |\vec{b}|^2=9 \).
3. Hence \( |\vec{b}|=3 \).

Teacher's Note:
1. Lagrange's identity combines the squares of the cross and dot products.
2. Take the positive square root for a magnitude.

Answer: (d) \( (\vec{a}×\vec{b})×\vec{c} \) is perpendicular to \( \vec{c} \)

Explanation:
1. \( (\vec{a}×\vec{b})×\vec{c} \) is a cross product with \( \vec{c} \) as the second factor.
2. So it is perpendicular to \( \vec{c} \).
3. The other statements are not true in general.

Teacher's Note:
1. A cross product is perpendicular to both of its factors.
2. Only \( \vec{c} \) and \( \vec{a}×\vec{b} \) are guaranteed to be perpendicular to it.

Answer: (c) \( 4(\hat{i}−\hat{k}) \)

Explanation:
1. The sum equals \( 2(\vec{a}×\vec{b}) \).
2. \( \vec{a}×\vec{b}=(1·1−1·(−1),\ 1·1−1·1,\ 1·(−1)−1·1)=(2,0,−2) \).
3. So the value is \( 2(2,0,−2)=4(\hat{i}−\hat{k}) \).

Teacher's Note:
1. Use the standard sum identity, then compute the cross product.
2. Do not forget the factor 2.

Answer: (a) \( 30^\circ \)

Explanation:
1. \( (\vec{a}·\vec{c})\vec{a}−(\vec{a}·\vec{a})\vec{c}+\vec{b}=\vec0 \), so \( \vec{b}=\vec{c}−(\vec{a}·\vec{c})\vec{a} \).
2. \( |\vec{b}|^2=|\vec{c}|^2−(\vec{a}·\vec{c})^2 \), so \( 1=4−(\vec{a}·\vec{c})^2 \) and \( (\vec{a}·\vec{c})^2=3 \).
3. With \( \vec{a}·\vec{c}=2\cosθ \), \( \cos^2θ=\frac34 \), so the acute angle is \( 30^\circ \).

Teacher's Note:
1. Isolate \( \vec{b} \) and take the magnitude of both sides.
2. Use \( \vec{a}·\vec{c}=|\vec{a}||\vec{c}|\cosθ \).

Answer: (c) \( \dfrac{\pi}{2} \)

Explanation:
1. By the Jacobi identity, \( \vec{a}×(\vec{b}×\vec{c})+\vec{b}×(\vec{c}×\vec{a})=−\vec{c}×(\vec{a}×\vec{b})=(\vec{c}·\vec{a})\vec{b}−(\vec{c}·\vec{b})\vec{a} \).
2. Equating this with \( \frac12\vec{a} \) gives \( \vec{c}·\vec{a}=0 \) and \( \vec{c}·\vec{b}=−\frac12 \) (since \( \vec{a},\vec{b} \) are independent).
3. So \( \vec{a} \) and \( \vec{c} \) are perpendicular.

Teacher's Note:
1. Use the cyclic identity to replace the sum of two triple products by the third with a minus sign.
2. Then compare coefficients.

Answer: (c) −1

Explanation:
1. \( (\vec{a}·\vec{c})\vec{b}−(\vec{a}·\vec{b})\vec{c}=\frac12\vec{b}+\frac12\vec{c} \).
2. So \( \vec{a}·\vec{c}=\frac12 \) and \( \vec{a}·\vec{b}=−\frac12 \), i.e. \( β=60^\circ \) and \( α=120^\circ \).
3. \( \cos(α+β)=\cos180^\circ=−1 \).

Teacher's Note:
1. Compare the coefficients of \( \vec{b} \) and \( \vec{c} \).
2. Convert the dot products to angles for unit vectors.

Answer: (a) \( \vec0 \)

Explanation:
1. Adding and subtracting the two conditions gives \( \vec{a}·\vec{c}=0 \) and \( \vec{b}·\vec{c}=0 \).
2. \( (\vec{a}×\vec{b})×\vec{c}=(\vec{a}·\vec{c})\vec{b}−(\vec{b}·\vec{c})\vec{a}=\vec0 \).

Teacher's Note:
1. First extract the two separate dot products.
2. Both coefficients in the triple product then vanish.