JEE Mathematics Vectors and their Applications MCQs Set 04

Answer: (c) \( \vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0 \)

Explanation:
1. Equality needs \( \vec{a}\times\vec{b} \) and \( \vec{c} \) to be parallel and \( \vec{a} \perp \vec{b} \), which in turn requires all three to be mutually perpendicular.

Teacher's Note:
1. Volume equals the product of edges only for a cuboid.

Answer: (c) \( \vec{\lambda}, \vec{\mu}, \vec{\nu} \) are coplanar

Explanation:
1. \( \vec{\lambda} + \vec{\mu} + \vec{\nu} = 0 \) because each pair of cross products cancels.
2. A zero linear combination means the three vectors are coplanar.

Teacher's Note:
1. Add the three vectors.
2. They sum to zero.

Answer: (c) \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \)

Explanation:
1. Both conditions give \( |\vec{a}||\vec{b}|\cos\theta = 0 \) and \( |\vec{a}||\vec{b}|\sin\theta = 0 \), which force \( |\vec{a}||\vec{b}| = 0 \).

Teacher's Note:
1. Both sin and cos cannot vanish.

Answer: (a) the null vector

Explanation:
1. The sum equals \( \vec{r}\times[(\vec{r}\cdot\hat{i})\hat{i} + (\vec{r}\cdot\hat{j})\hat{j} + (\vec{r}\cdot\hat{k})\hat{k}] = \vec{r}\times\vec{r} = \vec{0} \).

Teacher's Note:
1. Factor out r × ( ).

Answer: (d) \( 5\sqrt3 \)

Explanation:
1. Area \( = \frac12|\vec{d}_1\times\vec{d}_2| \). The cross product is \( -2\hat{i} - 14\hat{j} - 10\hat{k} \), magnitude \( \sqrt{300} = 10\sqrt3 \).
2. Area \( = 5\sqrt3 \).

Teacher's Note:
1. Half the cross product of diagonals.

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

Explanation:
1. \( \vec{a}\cdot\vec{b} \) is a scalar, and the cross product of a scalar with a vector is not defined.

Teacher's Note:
1. Check scalar versus vector at each step.

Answer: (a) \( \frac{(\vec{b}\times\vec{c})\times\vec{a}}{\vec{b}\cdot\vec{a}} \)

Explanation:
1. \( \vec{x} = \vec{c} + t\vec{b} \), and \( \vec{x}\cdot\vec{a} = 0 \) gives \( t = -\frac{\vec{c}\cdot\vec{a}}{\vec{b}\cdot\vec{a}} \).
2. \( \vec{x} = \frac{(\vec{b}\cdot\vec{a})\vec{c} - (\vec{c}\cdot\vec{a})\vec{b}}{\vec{b}\cdot\vec{a}} = \frac{(\vec{b}\times\vec{c})\times\vec{a}}{\vec{b}\cdot\vec{a}} \).

Teacher's Note:
1. x = c + tb, then use x·a = 0.

Answer: (b) \( \frac{1}{\sqrt2}(-\hat{j} + \hat{k}) \)

Explanation:
1. Write \( \vec{c} = \lambda\vec{a} + \mu\vec{b} \). \( \vec{c}\cdot\vec{a} = 6\lambda + 3\mu = 0 \) gives \( \mu = -2\lambda \).
2. \( \vec{c} = \lambda(\vec{a} - 2\vec{b}) = \lambda(-3\hat{j} + 3\hat{k}) \); the unit vector is \( \frac{1}{\sqrt2}(-\hat{j} + \hat{k}) \).

Teacher's Note:
1. Express c in terms of a and b.
2. Use c·a = 0.

Answer: (d) \( \mu, \frac\lambda2, \nu \) are in AP

Explanation:
1. Component equations: \( 2\lambda + \mu - 2\nu = 3 \), \( -\lambda + 3\mu + \nu = 2 \), \( \lambda - 2\mu - 3\nu = -5 \).
2. Adding the last two gives \( \mu - 2\nu = -3 \). Then the first gives \( \lambda = 3 \), so \( \nu = 2 \), \( \mu = 1 \).
3. \( \mu = 1, \frac\lambda2 = \frac32, \nu = 2 \) are in AP.

Teacher's Note:
1. Solve the three equations.
2. Test the AP condition.

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

Explanation:
1. \( (\vec{a}\times\vec{b})\times(\vec{a}\times\vec{c}) = [\vec{a}\,\vec{b}\,\vec{c}]\vec{a} \), and cyclically for the others.
2. Sum \( = [\vec{a}\,\vec{b}\,\vec{c}](\vec{a} + \vec{b} + \vec{c}) \).

Teacher's Note:
1. Each term reduces to [a b c] times a vector.

Answer: (b) \( 2\hat{i} + \hat{k} \)

Explanation:
1. Both sides have length \( \sqrt{11} \), so the equidistant point on BC is its midpoint.
2. \( \vec{AP} = \frac12(\vec{AB} + \vec{AC}) = 2\hat{i} + \hat{k} \).

Teacher's Note:
1. Equal sides make the bisector the median.

Answer: (a) \( 2\lambda \)

Explanation:
1. The edges \( \vec{a} + \vec{b} \), \( \vec{b} + \vec{c} \), \( \vec{c} + \vec{a} \) give \( [\vec{a} + \vec{b}\;\;\vec{b} + \vec{c}\;\;\vec{c} + \vec{a}] = 2\lambda \).

Teacher's Note:
1. Use the triple product of the face diagonals.

Answer: (a) \( \frac12|\vec{a} + \vec{b}| \)

Explanation:
1. \( |\vec{a} + \vec{b}|^2 = 2 + 2\cos\alpha = 4\cos^2\frac\alpha2 \).
2. So \( \cos\frac\alpha2 = \frac12|\vec{a} + \vec{b}| \).

Teacher's Note:
1. Use 1 + cos α = 2cos²(α/2).

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

Explanation:
1. The expression equals \( [\vec{a}\,\vec{b}\,\vec{c}]\vec{d} \) (resolution in the reciprocal basis), so its magnitude is \( |[\vec{a}\,\vec{b}\,\vec{c}]| \).

Teacher's Note:
1. Reciprocal basis resolution.

Answer: (b) 6

Explanation:
1. Area of OABC \( = \frac12|\vec{OB}\times\vec{AC}| = \frac12|(10\vec{a} + 2\vec{b})\times(\vec{b} - \vec{a})| = \frac12\cdot12|\vec{a}\times\vec{b}| = 6|\vec{a}\times\vec{b}| \).
2. \( q = |\vec{a}\times\vec{b}| \), so \( p/q = 6 \).

Teacher's Note:
1. Area of a quadrilateral from its diagonals.

Answer: (a) \( \vec{\mu}, \vec{\nu} \) are orthogonal vectors

Explanation:
1. \( \frac1a, \frac1b, \frac1c \) are in AP with common difference \( d \), so \( \frac1a = A + (p-1)d \) etc.
2. \( \vec{\mu}\cdot\vec{\nu} = (q-r)/a + (r-p)/b + (p-q)/c = d[(q-r)p + (r-p)q + (p-q)r] = 0 \). The vectors are orthogonal.

Teacher's Note:
1. Reciprocals of an HP are in AP.
2. Compute the dot product.

Answer: (a) 1

Explanation:
1. \( |\vec{a} + \vec{b} + \vec{c}|^2 = 3r^2 + 6\cdot\frac{r^2}2 = 6r^2 = 6 \).
2. So \( r = 1 \).

Teacher's Note:
1. Let the common magnitude be r.
2. Each dot product is r²/2.

Answer: (c) \( -3[\vec{a}\,\vec{b}\,\vec{c}] \)

Explanation:
1. The coefficient determinant \( \begin{vmatrix}1&1&1\\1&0&-1\\1&-1&0\end{vmatrix} = -3 \).

Teacher's Note:
1. Use the coefficient determinant.

Answer: (b) 0

Explanation:
1. The three vectors sum to zero, so they are coplanar and the triple product is 0.

Teacher's Note:
1. They add up to zero.

Answer: (c) \( \frac{4}{\sqrt3}(\hat{i} + \hat{j} + \hat{k}) \)

Explanation:
1. By symmetry the direction is \( \hat{i} + \hat{j} + \hat{k} \): its dot products with the three given vectors are all equal to 2.
2. The vector of magnitude 4 is \( \frac{4}{\sqrt3}(\hat{i} + \hat{j} + \hat{k}) \).

Teacher's Note:
1. Use symmetry.
2. Normalise to length 4.

Answer: (c) \( \vec{a} \perp \vec{b} \) and \( \vec{a} \parallel \vec{b}\times\vec{c} \)

Explanation:
1. \( \vec{c} = \vec{a}\times\vec{b} \) is perpendicular to \( \vec{a}, \vec{b} \) and \( \vec{b} = \vec{c}\times\vec{a} \) is perpendicular to \( \vec{a} \), so \( \vec{a}\perp\vec{b} \).
2. \( \vec{b}\times\vec{c} = \vec{b}\times(\vec{a}\times\vec{b}) = |\vec{b}|^2\vec{a} \), which is parallel to \( \vec{a} \).

Teacher's Note:
1. Check each claim with the properties of the cross product.

Answer: (b) \( \vec{b} + t\vec{a} \), where \( t \) is a scalar

Explanation:
1. \( (\vec{r} - \vec{b})\times\vec{a} = 0 \), so \( \vec{r} - \vec{b} \) is parallel to \( \vec{a} \).

Teacher's Note:
1. The difference is parallel to a.

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

Explanation:
1. The determinant \( \begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix} = 2 \), so the value is \( 2[\vec{a}\,\vec{b}\,\vec{c}] \).

Teacher's Note:
1. Use the coefficient determinant.

Answer: (a) \( -\frac1{\sqrt{26}}(4\hat{i} - \hat{j} - 3\hat{k}) \)

Explanation:
1. \( \vec{a}\times\vec{b} = 4\hat{i} - \hat{j} - 3\hat{k} \), magnitude \( \sqrt{26} \).
2. Its \( \hat{k} \) component is negative, so take the opposite direction: \( -\frac1{\sqrt{26}}(4\hat{i} - \hat{j} - 3\hat{k}) \).

Teacher's Note:
1. Cross product, then fix the sign.