JEE Mathematics Vectors and their Applications MCQs Set 05

Answer: (a) \( \cos^{-1}\frac45 \)

Explanation:
1. Adding: \( 3\vec{a} = 3\hat{i} - \hat{j} \), so \( \vec{a} = \hat{i} - \frac13\hat{j} \) and \( \vec{b} = \hat{i} + \frac13\hat{j} \).
2. \( \vec{a}\cdot\vec{b} = \frac89 \), \( |\vec{a}|^2 = \frac{10}9 \), so the cosine is \( \frac45 \).

Teacher's Note:
1. Solve for a and b.
2. The question asks the cosine, so the angle is cos inverse of the value.

Answer: (a) \( |\vec{u}| \)

Explanation:
1. \( |\vec{u}|^2 = 1 - 2\cos^2\theta + \cos^2\theta = \sin^2\theta \) and \( |\vec{v}| = \sin\theta \).

Teacher's Note:
1. Both equal sin θ.

Answer: (b) \( 4\vec{a}\cdot\vec{b} \)

Explanation:
1. Expanding, \( (|\vec{a}|^2 + 2\vec{a}\cdot\vec{b} + |\vec{b}|^2) - (|\vec{a}|^2 - 2\vec{a}\cdot\vec{b} + |\vec{b}|^2) = 4\vec{a}\cdot\vec{b} \).

Teacher's Note:
1. Expand both squares.

Answer: (d) \( \vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a} = \vec{0} \)

Explanation:
1. Collinear means \( (\vec{b} - \vec{a})\times(\vec{c} - \vec{a}) = \vec{0} \), which expands to \( \vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a} = \vec{0} \) (the vector \( (\vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a}) \) is twice the area vector).

Teacher's Note:
1. Use the area-vector condition.

Answer: (b) 0

Explanation:
1. The third vector equals \( \vec{a} \) plus the second, so the three are coplanar and the triple product is 0.

Teacher's Note:
1. Third = first + second.

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

Explanation:
1. \( |\vec{a} - \vec{b}|^2 = 2 - 2\cos\theta = 1 \), so \( \cos\theta = \frac12 \).
2. \( \theta = \frac\pi3 \).

Teacher's Note:
1. Expand
2. a - b
3. squared.
4. cos θ = 1/2.

Answer: (c) \( \frac12(\vec{p} + \vec{q} + \vec{r}) \)

Explanation:
1. With common magnitude \( m \), \( \vec{p}\times((\vec{x} - \vec{q})\times\vec{p}) = m^2(\vec{x} - \vec{q}) - (\vec{p}\cdot\vec{x})\vec{p} \).
2. Summing: \( 3m^2\vec{x} - m^2(\vec{p} + \vec{q} + \vec{r}) - m^2\vec{x} = 0 \), so \( \vec{x} = \frac12(\vec{p} + \vec{q} + \vec{r}) \).

Teacher's Note:
1. Expand each triple product.

Answer: (a) 16

Explanation:
1. \( [\vec{a}\times\vec{b}\;\;\vec{b}\times\vec{c}\;\;\vec{c}\times\vec{a}] = [\vec{a}\,\vec{b}\,\vec{c}]^2 = 16 \).

Teacher's Note:
1. Square of the triple product.

Answer: (b) along the line of intersection of the plane containing \( \vec{a}, \vec{b} \) and the plane containing \( \vec{c}, \vec{d} \)

Explanation:
1. It is perpendicular to both normals \( \vec{a}\times\vec{b} \) and \( \vec{c}\times\vec{d} \), so it lies in both planes.

Teacher's Note:
1. Perpendicular to both plane normals.

Answer: (c) \( 2\vec{a}^2 \)

Explanation:
1. \( |\vec{a}\times\hat{i}|^2 = |\vec{a}|^2 - (\vec{a}\cdot\hat{i})^2 \). Summing: \( 3|\vec{a}|^2 - |\vec{a}|^2 = 2|\vec{a}|^2 \).

Teacher's Note:
1. Use Lagrange's identity.

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

Explanation:
1. In a regular hexagon \( \vec{OC} = \vec{OB} - \vec{OA} \), because \( \vec{OA} + \vec{OC} = \vec{OB} \).
2. So \( \vec{BC} = \vec{OC} - \vec{OB} = -\vec{OA} = -\hat{i} + \hat{j} - 2\hat{k} \).

Teacher's Note:
1. Use the hexagon symmetry.
2. OC = OB - OA.

Answer: (c) \( 2[\vec{a}\,\vec{b}\,\vec{c}]\vec{r} \)

Explanation:
1. Each term is \( [\vec{a}\,\vec{b}\,\vec{c}]\vec{r} \) minus a multiple of \( \vec{a}, \vec{b} \) or \( \vec{c} \), e.g. \( [\vec{a}\,\vec{b}\,\vec{c}]\vec{r} - [\vec{a}\,\vec{b}\,\vec{r}]\vec{c} \).
2. Sum \( = 3[\vec{a}\,\vec{b}\,\vec{c}]\vec{r} - [\vec{a}\,\vec{b}\,\vec{c}]\vec{r} = 2[\vec{a}\,\vec{b}\,\vec{c}]\vec{r} \).

Teacher's Note:
1. Expand with the vector triple product.

Answer: (b) \( \hat{i} \)

Explanation:
1. The first gives \( a_1 = 1 \); the second gives \( a_1 + a_2 = 1 \) so \( a_2 = 0 \); the third gives \( a_3 = 0 \).
2. So \( \vec{a} = \hat{i} \).

Teacher's Note:
1. Subtract successive equations.

Answer: (b) \( \vec{a}\times\vec{b} \)

Explanation:
1. \( [\vec{a}\,\vec{b}\,\hat{i}] = (\vec{a}\times\vec{b})\cdot\hat{i} \) is the \( x \) component of \( \vec{a}\times\vec{b} \); the sum rebuilds the vector \( \vec{a}\times\vec{b} \).

Teacher's Note:
1. These are the components of a×b.

Answer: (a) \( |\vec{a}|^2 \)

Explanation:
1. The sum of squares of the components is \( |\vec{a}|^2 \).

Teacher's Note:
1. Sum of squared components is the squared length.

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

Explanation:
1. The sum is the diagonal of a rhombus, which bisects the \( 60^\circ \) angle.
2. So the angle is \( 30^\circ \).

Teacher's Note:
1. Equal magnitudes make a rhombus.
2. Diagonal bisects the angle.

Answer: (c) \( \frac2{\sqrt{10}}(3\hat{i} - \hat{k}) \)

Explanation:
1. Both vectors have length 3, so the internal bisector is along their sum \( 3\hat{i} - \hat{k} \), of length \( \sqrt{10} \).
2. Scaling to magnitude 2 gives \( \frac2{\sqrt{10}}(3\hat{i} - \hat{k}) \).

Teacher's Note:
1. Equal lengths make the sum a bisector.

Answer: (c) 3

Explanation:
1. Reciprocal system: \( \vec{a}\cdot\vec{p} = \vec{b}\cdot\vec{q} = \vec{c}\cdot\vec{r} = 1 \), while the mixed products vanish.
2. So the sum is \( 1 + 1 + 1 = 3 \).

Teacher's Note:
1. Use the reciprocal system.

Answer: (b) -1

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

Teacher's Note:
1. Expand the squared magnitude.
2. a·b vanishes.

Answer: (c) 0

Explanation:
1. Coplanarity gives \( \begin{vmatrix}1&-2&3\\2&3&-1\\\lambda&1&2\lambda-1\end{vmatrix} = 7\lambda = 0 \).
2. So \( \lambda = 0 \).

Teacher's Note:
1. Set the scalar triple product to zero.

Answer: (c) 0

Explanation:
1. \( \vec{r}\cdot(\vec{a} + \vec{b} + \vec{c}) = (l + m + n)[\vec{a}\,\vec{b}\,\vec{c}] = 0 \), so \( l + m + n = 0 \).

Teacher's Note:
1. Dot with a+b+c.

Answer: (b) \( \vec{a} \perp \vec{b} \)

Explanation:
1. Squaring, \( 4\vec{a}\cdot\vec{b} = 0 \), so the vectors are perpendicular.

Teacher's Note:
1. The diagonals of a parallelogram are equal.

Answer: (d) none of these

Explanation:
1. Third vertex \( = 3(\text{centroid}) - (\text{sum of the other two}) = 3\hat{k} - (3\hat{i} + 4\hat{k}) = -3\hat{i} - \hat{k} \).
2. This matches none of (a), (b), (c).

Teacher's Note:
1. Centroid is the mean of the vertices.
2. Third = 3G minus the other two.

Answer: (c) \( \frac{|\vec{b}\times\vec{c}|}{|\vec{b} - \vec{c}|} \)

Explanation:
1. Area of the triangle \( = \frac12|\vec{b}\times\vec{c}| = \frac12\cdot BC\cdot h \), with \( BC = |\vec{b} - \vec{c}| \).

Teacher's Note:
1. Compute the area two ways.