JEE Mathematics Vectors MCQs Set 29

Answer: (d) \( 5\sqrt{2} \)

Explanation:
1. The three conditions give \( \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = 0 \), \( \vec{b}\cdot\vec{c} + \vec{b}\cdot\vec{a} = 0 \) and \( \vec{c}\cdot\vec{a} + \vec{c}\cdot\vec{b} = 0 \).
2. Adding and solving, each of the three dot products is 0.
3. \( |\vec{a} + \vec{b} + \vec{c}|^2 = 9 + 16 + 25 = 50 \), so the magnitude is \( 5\sqrt{2} \).

Teacher's Note:
1. Write each perpendicularity as a dot product equal to zero.
2. The three equations force all pairwise dot products to vanish.

Answer: (c) \( k\sqrt{4 + 2\sqrt{2}} \)

Explanation:
1. Write the forces as \( k\hat{i} \), \( \frac{k}{\sqrt{2}}(\hat{i} + \hat{j}) \), \( k\hat{j} \) and \( \frac{k}{\sqrt{2}}(-\hat{i} + \hat{j}) \).
2. The sum is \( k\hat{i} + (k + \sqrt{2}k)\hat{j} \).
3. Magnitude \( = k\sqrt{1 + (1 + \sqrt{2})^2} = k\sqrt{4 + 2\sqrt{2}} \).

Teacher's Note:
1. Resolve each force along the two axes and add the components.
2. Use \( (1 + \sqrt{2})^2 = 3 + 2\sqrt{2} \).

Answer: (c) \( \vec{XN} = \frac{3}{4}\vec{MN} \)

Explanation:
1. Take A as origin with \( \vec{AB} = \vec{b} \), \( \vec{AC} = \vec{c} \). Then \( M = \frac{\vec{c}}{3} \) and \( N = 2\vec{b} - \vec{c} \).
2. Points of MN: \( \frac{\vec{c}}{3} + t\left(2\vec{b} - \frac{4\vec{c}}{3}\right) \). It meets line AB when the \( \vec{c} \)-coefficient is zero: \( t = \frac{1}{4} \).
3. So \( X = \frac{\vec{b}}{2} \), and \( \vec{MX} = \frac{1}{4}\vec{MN} \), i.e. \( \vec{XN} = \frac{3}{4}\vec{MN} \).

Teacher's Note:
1. Use non-collinear vectors \( \vec{b}, \vec{c} \) as a basis.
2. Coefficients of independent vectors must match separately.

Answer: (a) 6

Explanation:
1. \( \vec{a}\times(\vec{b} - 2\vec{c}) = \vec{0} \), so \( \vec{b} - 2\vec{c} = \alpha\vec{a} \).
2. Squaring: \( 16 + 4 - 4\vec{b}\cdot\vec{c} = \alpha^2 \), and \( \vec{b}\cdot\vec{c} = 4\cdot 1\cdot\frac{1}{4} = 1 \), so \( \alpha^2 = 16 \), \( \alpha = \pm 4 \).
3. Then \( \vec{b} = 2\vec{c} \pm 4\vec{a} \), so \( |\ell| + |\mu| = 2 + 4 = 6 \).

Teacher's Note:
1. Parallel vectors have zero cross product.
2. Square the relation to bring in the given angle.

Answer: (c) \( [0, \frac{2\pi}{3}] \)

Explanation:
1. \( \cos\theta = \frac{ab + bc + ca}{a^2 + b^2 + c^2} \).
2. Since \( (a - b)^2 + (b - c)^2 + (c - a)^2 \ge 0 \), \( \cos\theta \le 1 \); and \( (a + b + c)^2 \ge 0 \) gives \( \cos\theta \ge -\frac{1}{2} \).
3. So \( -\frac{1}{2} \le \cos\theta \le 1 \), i.e. \( 0 \le \theta \le \frac{2\pi}{3} \).

Teacher's Note:
1. Both bounds come from sums of squares.
2. \( \cos\theta = -\frac{1}{2} \) corresponds to \( \frac{2\pi}{3} \).

Answer: (d) \( [\vec{a}\ \vec{b}\ \vec{c}] = |\vec{a}|^2 \)

Explanation:
1. \( \vec{a}\cdot\vec{b} = 0 \) gives \( x - y + 2 = 0 \); \( \vec{a}\cdot\vec{c} = 4 \) gives \( x + 2y = 4 \). So \( x = 0,\ y = 2 \).
2. \( \vec{a} = 2\hat{j} + 2\hat{k} \) and \( |\vec{a}|^2 = 8 \).
3. \( [\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} 0 & 2 & 2 \\ 1 & -1 & 1 \\ 1 & 2 & 0 \end{vmatrix} = 8 = |\vec{a}|^2 \).

Teacher's Note:
1. Solve the two linear equations for \( x \) and \( y \) first.
2. Evaluate the determinant to get the scalar triple product.

Answer: (b) coplanar but cannot form a triangle

Explanation:
1. \( \vec{v}_1 + \vec{v}_2 = 3\hat{i} + \hat{j} + 4\hat{k} = \vec{v}_3 \), so the vectors are linearly dependent and hence coplanar.
2. For a closed triangle we would need \( \vec{v}_1 + \vec{v}_2 + \vec{v}_3 = \vec{0} \), but the sum is \( 2\vec{v}_3 \ne \vec{0} \).
3. So they are coplanar but cannot form a triangle.

Teacher's Note:
1. Check whether one vector is the sum of the other two.
2. A triangle needs the three vectors to sum to zero.

Answer: (c) \( (1 - \cos\theta)\sqrt{1 + 2\cos\theta} \)

Explanation:
1. \( [\vec{a}\ \vec{b}\ \vec{c}]^2 = \begin{vmatrix} 1 & \cos\theta & \cos\theta \\ \cos\theta & 1 & \cos\theta \\ \cos\theta & \cos\theta & 1 \end{vmatrix} \).
2. With \( c = \cos\theta \) the determinant is \( 1 - 3c^2 + 2c^3 = (1 - c)^2(1 + 2c) \).
3. So \( |[\vec{a}\ \vec{b}\ \vec{c}]| = (1 - \cos\theta)\sqrt{1 + 2\cos\theta} \).

Teacher's Note:
1. The square of a scalar triple product is a Gram determinant.
2. Factorise \( 2c^3 - 3c^2 + 1 \) using the root \( c = 1 \).

Answer: (a) \( \vec{a}_1 + \vec{a}_2 + \vec{a}_3 + \vec{a}_4 = \vec{0} \)

Explanation:
1. Take vertices A, B, C, D. The outward area vectors are \( \frac{1}{2}\vec{AC}\times\vec{AB} \), \( \frac{1}{2}\vec{DB}\times\vec{DC} \), \( \frac{1}{2}\vec{DC}\times\vec{DA} \), \( \frac{1}{2}\vec{DA}\times\vec{DB} \).
2. With D as origin these are \( \frac{1}{2}[(\vec{c}-\vec{a})\times(\vec{b}-\vec{a})] \), \( \frac{1}{2}\vec{b}\times\vec{c} \), \( \frac{1}{2}\vec{c}\times\vec{a} \), \( \frac{1}{2}\vec{a}\times\vec{b} \).
3. Expanding the first gives \( \frac{1}{2}(\vec{c}\times\vec{b} - \vec{c}\times\vec{a} - \vec{a}\times\vec{b}) \), and the four sum to zero.

Teacher's Note:
1. The vector area of a closed surface is zero.
2. Place one vertex at the origin to simplify the cross products.

Answer: (c) 0

Explanation:
1. \( \vec{r}\cdot(\vec{a} + \vec{b} + \vec{c}) = 0 \).
2. Each cross product is perpendicular to two of the vectors: \( \ell[\vec{a}\ \vec{b}\ \vec{c}] + m[\vec{a}\ \vec{b}\ \vec{c}] + n[\vec{a}\ \vec{b}\ \vec{c}] = 0 \).
3. Since the triple product is 2, not zero, \( \ell + m + n = 0 \).

Teacher's Note:
1. Expand the dot product term by term.
2. Triple products with a repeated vector vanish.

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

Explanation:
1. Write \( \vec{r} = x\vec{a} + y\vec{b} + z\vec{c} \). Then \( (\vec{a}\times\vec{b})\times(\vec{r}\times\vec{c}) = [\vec{a}\ \vec{b}\ \vec{c}]\vec{r} - [\vec{a}\ \vec{b}\ \vec{r}]\vec{c} \) and \( [\vec{a}\ \vec{b}\ \vec{r}] = z[\vec{a}\ \vec{b}\ \vec{c}] \).
2. So the first term is \( [\vec{a}\ \vec{b}\ \vec{c}](\vec{r} - z\vec{c}) \); similarly for the others.
3. The sum is \( [\vec{a}\ \vec{b}\ \vec{c}](3\vec{r} - x\vec{a} - y\vec{b} - z\vec{c}) = 2[\vec{a}\ \vec{b}\ \vec{c}]\vec{r} \).

Teacher's Note:
1. Expand each vector product with the identity for \( (\vec{p}\times\vec{q})\times(\vec{r}\times\vec{s}) \).
2. Resolve \( \vec{r} \) along the three non-coplanar vectors.

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

Explanation:
1. \( (\vec{a}\times\vec{b})\cdot(\vec{a}\times\vec{c}) = (\vec{a}\cdot\vec{a})(\vec{b}\cdot\vec{c}) - (\vec{a}\cdot\vec{c})(\vec{b}\cdot\vec{a}) \).
2. Since \( \vec{a} \) is parallel to \( \vec{b}\times\vec{c} \), it is perpendicular to both \( \vec{b} \) and \( \vec{c} \), so \( \vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c} = 0 \).
3. The result is \( a^2(\vec{b}\cdot\vec{c}) \).

Teacher's Note:
1. Use the Lagrange identity for the dot product of two cross products.
2. A vector parallel to \( \vec{b}\times\vec{c} \) is perpendicular to \( \vec{b} \) and \( \vec{c} \).

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

Explanation:
1. Let \( \vec{b} = (b_1, b_2, b_3) \). \( \vec{a}\times\vec{b} = (b_3 + b_2,\ -b_1,\ -b_1) \) must equal \( -\vec{c} = (-1, 1, 1) \).
2. So \( b_1 = -1 \) and \( b_2 + b_3 = -1 \). Also \( \vec{a}\cdot\vec{b} = b_2 - b_3 = 3 \).
3. Solving gives \( b_2 = 1,\ b_3 = -2 \): \( \vec{b} = -\hat{i} + \hat{j} - 2\hat{k} \).

Teacher's Note:
1. Expand the cross product component by component.
2. Use the dot-product condition to separate \( b_2 \) and \( b_3 \).

Answer: (d) \( (-3, 2) \)

Explanation:
1. \( \vec{a}\cdot\vec{c} = \lambda - 1 + 2\mu = 0 \).
2. \( \vec{b}\cdot\vec{c} = 2\lambda + 4 + \mu = 0 \).
3. Solving the two equations: \( \lambda = -3,\ \mu = 2 \).

Teacher's Note:
1. Mutually orthogonal means every pair has zero dot product.
2. \( \vec{a}\cdot\vec{b} = 2 - 4 + 2 = 0 \) already holds.

Answer: (a) \( -5 \)

Explanation:
1. Both are unit vectors and \( \vec{a}\cdot\vec{b} = \frac{6 - 6}{7\sqrt{10}} = 0 \).
2. \( (\vec{a}\times\vec{b})\times\vec{a} = \vec{b}(\vec{a}\cdot\vec{a}) - \vec{a}(\vec{a}\cdot\vec{b}) = \vec{b} \) and \( (\vec{a}\times\vec{b})\times\vec{b} = \vec{b}(\vec{a}\cdot\vec{b}) - \vec{a}(\vec{b}\cdot\vec{b}) = -\vec{a} \), so \( (\vec{a}\times\vec{b})\times(\vec{a} + 2\vec{b}) = \vec{b} - 2\vec{a} \).
3. So the expression is \( (2\vec{a} - \vec{b})\cdot(\vec{b} - 2\vec{a}) = -|2\vec{a} - \vec{b}|^2 = -(4 + 1) = -5 \).

Teacher's Note:
1. Check whether the vectors are unit and orthogonal first.
2. For orthonormal \( \vec{a}, \vec{b} \), \( \vec{a}\times\vec{b} \) is a third unit vector.

Answer: (d) \( -2 \)

Explanation:
1. Coplanar means \( \begin{vmatrix} p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & r \end{vmatrix} = 0 \).
2. Expanding: \( pqr - p - q - r + 2 = 0 \).
3. So \( pqr - (p + q + r) = -2 \).

Teacher's Note:
1. Three vectors are coplanar when their determinant is zero.
2. Expand along the first row.

Answer: (c) \( \vec{0} \)

Explanation:
1. Let \( \vec{a} + 3\vec{b} = \lambda\vec{c} \) and \( \vec{b} + 2\vec{c} = \mu\vec{a} \).
2. From the second, \( 3\vec{b} = 3\mu\vec{a} - 6\vec{c} \). Substitute into the first: \( (1 + 3\mu)\vec{a} = (\lambda + 6)\vec{c} \).
3. Since \( \vec{a}, \vec{c} \) are not collinear, \( \lambda = -6 \), so \( \vec{a} + 3\vec{b} + 6\vec{c} = \vec{0} \).

Teacher's Note:
1. Eliminate \( \vec{b} \) between the two relations.
2. Non-collinear vectors force both coefficients to vanish.

Answer: (c) \( \frac{\pi}{3} \)

Explanation:
1. \( \vec{c}\cdot\vec{d} = (\hat{a} + 2\hat{b})\cdot(5\hat{a} - 4\hat{b}) = 0 \).
2. \( 5 - 4\hat{a}\cdot\hat{b} + 10\hat{a}\cdot\hat{b} - 8 = 0 \), so \( 6\hat{a}\cdot\hat{b} = 3 \).
3. \( \cos\theta = \frac{1}{2} \), so \( \theta = \frac{\pi}{3} \).

Teacher's Note:
1. Expand the dot product using \( \hat{a}\cdot\hat{a} = \hat{b}\cdot\hat{b} = 1 \).
2. Perpendicular vectors have zero dot product.

Answer: (c) \( \sqrt{\frac{3}{2}} \)

Explanation:
1. A normal to the plane is \( \vec{n} = (\hat{i} + \hat{j} + \hat{k})\times(\hat{i} + 2\hat{j} + 3\hat{k}) = \hat{i} - 2\hat{j} + \hat{k} \).
2. Projection magnitude \( = \frac{|(2\hat{i} + 3\hat{j} + \hat{k})\cdot\vec{n}|}{|\vec{n}|} = \frac{|2 - 6 + 1|}{\sqrt{6}} \).
3. This equals \( \frac{3}{\sqrt{6}} = \sqrt{\frac{3}{2}} \).

Teacher's Note:
1. The cross product gives a vector perpendicular to the plane.
2. Projection of \( \vec{u} \) on \( \vec{n} \) is \( \frac{\vec{u}\cdot\vec{n}}{|\vec{n}|} \).

Answer: (c) \( \frac{1}{\sqrt{3}} \)

Explanation:
1. \( V = \begin{vmatrix} 1 & \lambda & 1 \\ 0 & 1 & \lambda \\ \lambda & 0 & 1 \end{vmatrix} = \lambda^3 - \lambda + 1 \).
2. \( \frac{dV}{d\lambda} = 3\lambda^2 - 1 = 0 \) gives \( \lambda = \pm\frac{1}{\sqrt{3}} \).
3. \( \frac{d^2V}{d\lambda^2} = 6\lambda > 0 \) only for \( \lambda = \frac{1}{\sqrt{3}} \), which gives the minimum.

Teacher's Note:
1. Volume is the scalar triple product.
2. Use the second derivative to separate the minimum from the maximum.