JEE Mathematics Vectors MCQs Set 23

Answer: (d) \( \hat i \)

Explanation:
1. \( \vec a\times(\vec a\times\vec b)=\vec a(\vec a\cdot\vec b)-\vec b|\vec a|^2=\vec a-3\vec b \).
2. Also \( \vec a\times(\hat j-\hat k)=-2\hat i+\hat j+\hat k \).
3. So \( 3\vec b=\vec a+2\hat i-\hat j-\hat k=3\hat i \) and \( \vec b=\hat i \). Check: \( \vec a\cdot\hat i=1 \) and \( \vec a\times\hat i=\hat j-\hat k \).

Teacher's Note:
1. Cross the given cross product with a again to isolate b.
2. Verify against both given conditions.

Answer: (a) \( -(\vec a+\vec b) \)

Explanation:
1. In an equilateral triangle the orthocentre coincides with the centroid, so \( \vec{OA}+\vec{OB}+\vec{OC}=\vec0 \).
2. Hence \( \vec{OC}=-(\vec a+\vec b) \).

Teacher's Note:
1. Centroid, orthocentre, circumcentre and incentre coincide for an equilateral triangle.
2. The position vectors of the vertices sum to zero about the centroid.

Answer: (c) 1

Explanation:
1. Let \( \vec n=\vec a\times\vec b \). Then \( [\vec c\,\vec a\,\vec b]=\vec c\cdot\vec n \).
2. \( (\vec n-\vec a\times\vec n)\cdot\vec n=|\vec n|^2-0 \) because \( \vec a\times\vec n\perp\vec n \).
3. So the value is \( \frac{|\vec n|^2}{|\vec n|^2}=1 \).

Teacher's Note:
1. Reduce the expression to a standard form before reading off the answer.
2. Verify with a simple check of the final result.

Answer: (c) parallelogram, which is neither a rhombus nor a rectangle

Explanation:
1. \( \vec{PQ}=(6,1)=\vec{SR} \), so it is a parallelogram.
2. \( \vec{PS}=(-1,3) \): \( \vec{PQ}\cdot\vec{PS}=-3\ne0 \) (not a rectangle) and \( |PQ|^2=37\ne10=|PS|^2 \) (not a rhombus).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \frac\pi4\ \text{or}\ \frac{3\pi}4 \)

Explanation:
1. Normal of \( P_1 \): \( (2\hat j+3\hat k)\times(4\hat j-3\hat k)=-18\hat i \) (plane \( x=0 \)). Normal of \( P_2 \): \( (\hat j-\hat k)\times(3\hat i+3\hat j)=3(\hat i-\hat j-\hat k) \).
2. So \( \vec A\parallel(0,1,-1) \).
3. Then \( \cos\theta=\frac{\pm3}{\sqrt2\cdot3}=\pm\frac1{\sqrt2} \), so the angle is \( \frac\pi4 \) or \( \frac{3\pi}4 \) depending on the direction of \( \vec A \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) neither x nor y

Explanation:
1. Expanding the determinant \( \begin{vmatrix}1&0&-1\\x&1&1-x\\y&x&1+x-y\end{vmatrix} \) gives the constant 1.
2. So it depends on neither \( x \) nor \( y \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( \frac1{\sqrt2} \)

Explanation:
1. \( V^2 \) equals the Gram determinant \( \begin{vmatrix}1&\frac12&\frac12\\\frac12&1&\frac12\\\frac12&\frac12&1\end{vmatrix}=\frac34-\frac18-\frac18=\frac12 \).
2. So \( V=\frac1{\sqrt2} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) 0

Explanation:
1. The three vectors \( 2\vec a-\vec b,\ 2\vec b-\vec c,\ 2\vec c-\vec a \) are linear combinations of coplanar vectors, so they are themselves coplanar.
2. The scalar triple product of coplanar vectors is 0.

Teacher's Note:
1. Combinations of vectors in a plane remain in that plane.
2. Coplanar vectors have zero triple product.

Answer: (d) 2

Explanation:
1. The determinant is \( \tan A\tan B\tan C-(\tan A+\tan B+\tan C)+2 \).
2. In a triangle \( \tan A+\tan B+\tan C=\tan A\tan B\tan C \), so the value is 2.

Teacher's Note:
1. Work from the given condition to find the function or quantity, then substitute.
2. Verify the answer on a simple case.

Answer: (a) 0

Explanation:
1. \( \vec a\times\vec b \) and \( \vec c\times\vec d \) are normals to \( P_1 \) and \( P_2 \).
2. Their cross product is zero, so the normals are parallel: the planes are parallel and the angle between them is 0.

Teacher's Note:
1. The angle between planes is the angle between their normals.
2. Zero cross product means parallel vectors.

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 dot-product sum is at least \( -\frac32 \).
3. So the expression is at most \( 6+3=9 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) Statement 1 is true and Statement 2 is false

Explanation:
1. In a regular hexagon \( \vec{ST} \) is antiparallel to \( \vec{PQ} \), so \( \vec{PQ}\times\vec{ST}=\vec0 \) and Statement 2 is false.
2. Also \( \vec{RS} \) makes an angle of 120 degrees with \( \vec{PQ} \), so \( \vec{PQ}\times\vec{RS}\ne\vec0 \).
3. Statement 1: \( \vec{RS}+\vec{ST}=\vec{RT} \) is not parallel to \( \vec{PQ} \), so the cross product is non-zero: true.

Teacher's Note:
1. Directions of consecutive sides of a regular hexagon differ by 60 degrees.
2. Opposite sides are parallel.

Answer: (c) \( \sqrt{59} \)

Explanation:
1. \( [\vec U\,\vec V\,\vec W]=\vec U\cdot(\vec V\times\vec W) \) and \( \vec V\times\vec W=(3,-7,-1) \).
2. The maximum for a unit \( \vec U \) is \( |\vec V\times\vec W|=\sqrt{9+49+1}=\sqrt{59} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. Going round the triangle, \( \vec a+\vec b+\vec c=\vec0 \).
2. Cross with \( \vec a \): \( \vec a\times\vec b=-\vec a\times\vec c=\vec c\times\vec a \); cross with \( \vec c \) to get \( \vec b\times\vec c=\vec a\times\vec b \).
3. So the three cross products are equal (their sum is \( 3\,\vec a\times\vec b\ne0 \)).

Teacher's Note:
1. Vectors along the sides of a closed triangle sum to zero.
2. Cross the relation with each vector in turn.

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

Explanation:
1. \( |\vec a+\vec b|^2=3|\vec c|^2=3 \), so \( 2+2\vec a\cdot\vec b=3 \) and \( \cos\theta=\frac12 \), giving \( \theta=\frac\pi3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \vec b,\vec d \) are non-parallel

Explanation:
1. Since \( |\vec a\times\vec b|\le1 \) and \( |\vec c\times\vec d|\le1 \), the product 1 forces both to be unit vectors along the same direction \( \hat n \).
2. Then \( \vec c\perp\hat n \) and \( \vec b\perp\hat n \), so \( \vec a,\vec b,\vec c \) are coplanar and \( \vec b,\vec c,\vec d \) are coplanar.
3. If \( \vec b\parallel\vec d \), \( (\vec a\cdot\vec c)(\vec b\cdot\vec d)-(\vec a\cdot\vec d)(\vec b\cdot\vec c) \) cannot equal 1 with \( \vec a\cdot\vec c=\frac12 \), so \( \vec b,\vec d \) are non-parallel.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \hat u=\frac{\hat a+\hat b}{|\hat a+\hat b|},\ M=(1+\hat a\cdot\hat b)^{1/2} \)

Explanation:
1. \( |\vec{OP}|^2=\cos^2t+\sin^2t+2\hat a\cdot\hat b\sin t\cos t=1+\hat a\cdot\hat b\sin2t \).
2. It is largest when \( \sin2t=1 \), i.e. \( t=\frac\pi4 \), giving \( M=\sqrt{1+\hat a\cdot\hat b} \) and \( \vec{OP}=\frac{\hat a+\hat b}{\sqrt2} \) along \( \hat a+\hat b \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( 60^\circ \)

Explanation:
1. \( (\vec a+2\vec b)\cdot(5\vec a-4\vec b)=5-4\,\vec a\cdot\vec b+10\,\vec a\cdot\vec b-8=0 \).
2. So \( 6\,\vec a\cdot\vec b=3 \), \( \cos\theta=\frac12 \) and \( \theta=60^\circ \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. A vector in the plane of \( \vec b,\vec c \) is \( \lambda\vec b+\mu\vec c=(2\lambda+\mu,\ \lambda-\mu,\ \lambda+\mu) \).
2. Orthogonal to \( \vec a \): \( 3(2\lambda+\mu)+2(\lambda-\mu)+6(\lambda+\mu)=14\lambda+7\mu=0 \), so \( \mu=-2\lambda \).
3. The vector is proportional to \( 3\hat j-\hat k \) with length \( \sqrt{10} \).

Teacher's Note:
1. Write the coplanar vector as a combination of b and c.
2. Apply the orthogonality condition and normalise.

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

Explanation:
1. Cross the relation with \( \vec a \): \( \vec a\times\vec b+\vec a\times\vec c=\vec0 \), so \( \vec a\times\vec b=\vec c\times\vec a \).
2. Cross with \( \vec c \) from the right: \( \vec a\times\vec c+\vec b\times\vec c=\vec0 \), giving \( \vec a\times\vec b=\vec b\times\vec c \).
3. So (i) and (iii) hold; (iv) is the negative.

Teacher's Note:
1. When three vectors sum to zero, all three pairwise cross products are equal.
2. Reversing the order of a cross product flips its sign.

Answer: (d) two

Explanation:
1. Coplanar means \( \begin{vmatrix}-\lambda^2&1&1\\1&-\lambda^2&1\\1&1&-\lambda^2\end{vmatrix}=0 \).
2. The determinant equals \( (\lambda^2+1)^2(2-\lambda^2) \), which vanishes only for \( \lambda^2=2 \).
3. So \( \lambda=\pm\sqrt2 \): two real values.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \pi \)

Explanation:
1. Volume \( =\left|\begin{vmatrix}1&1&0\\1&2&0\\1&1&\pi\end{vmatrix}\right|=\pi(2-1)=\pi \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \sqrt{4+\sqrt3+\sqrt2} \)

Explanation:
1. \( |\vec a+\vec b+\vec c|^2=3+2(\cos30^\circ+\cos60^\circ+\cos45^\circ) \).
2. This equals \( 3+\sqrt3+1+\sqrt2=4+\sqrt3+\sqrt2 \).

Teacher's Note:
1. Expand the square of the sum using dot products.
2. For unit vectors the dot product equals the cosine of the angle.

Answer: (a) 5

Explanation:
1. \( \vec a\cdot\vec b=\frac{2-2}{\sqrt{70}}=0 \) and both are unit vectors.
2. \( (\vec a\times\vec b)\times(\vec a-2\vec b)=\vec b(1-2\vec a\cdot\vec b)+\vec a(2-\vec a\cdot\vec b)=\vec b+2\vec a \).
3. So the value is \( |2\vec a+\vec b|^2=4+1+0=5 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( 4\hat i-\hat j+4\hat k \)

Explanation:
1. \( \lambda\vec a+\mu\vec b=(\lambda+\mu,\ 2\lambda-\mu,\ \lambda+\mu) \): the first and third components are equal, which only \( 4\hat i-\hat j+4\hat k \) among the options satisfies (\( \lambda=1,\ \mu=3 \)).
2. Projection on \( \vec c \): \( \frac{4-1-4}{\sqrt3}=-\frac1{\sqrt3} \), of magnitude \( \frac1{\sqrt3} \).

Teacher's Note:
1. A vector in the plane of a and b has equal x and z components here.
2. Projection = dot product divided by |c|.

Answer: (a) \( -\vec b \)

Explanation:
1. Use \( \vec a\times(\vec a\times\vec b)=\vec a(\vec a\cdot\vec b)-\vec b(\vec a\cdot\vec a) \).
2. Here \( \vec a\cdot\vec b=0 \) and \( \vec a\cdot\vec a=1 \), so the result is \( -\vec b \).

Teacher's Note:
1. Apply the vector triple product expansion.
2. The result lies in the plane of a and b, perpendicular to a.