JEE Mathematics Vectors and their Applications MCQs Set 06

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

Explanation:
1. \( [\vec{b}\,\vec{c}\,\vec{b}\times\vec{c}] = |\vec{b}\times\vec{c}|^2 \), and \( |\vec{b}\times\vec{c}|^2 + (\vec{b}\cdot\vec{c})^2 = |\vec{b}|^2|\vec{c}|^2 \).

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

Answer: (a) \( \sqrt6 \)

Explanation:
1. Adding the three orthogonality conditions gives \( \vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0 \).
2. \( |\vec{a} + \vec{b} + \vec{c}|^2 = 1 + 2 + 3 = 6 \).

Teacher's Note:
1. All pairwise dot products vanish.
2. Sum the squares.

Answer: (a) \( \frac32 \)

Explanation:
1. \( \vec{a}\times\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k} \), magnitude 3. \( |\vec{c} - \vec{a}|^2 = |\vec{c}|^2 - 2|\vec{c}| + 9 = 8 \) gives \( |\vec{c}| = 1 \).
2. \( |(\vec{a}\times\vec{b})\times\vec{c}| = 3\cdot1\cdot\sin30^\circ = \frac32 \).

Teacher's Note:
1. Find
2. c
3. first.

Answer: (a) 3

Explanation:
1. Each numerator and denominator equals \( [\vec{a}\,\vec{b}\,\vec{c}] = 1 \), so each fraction is 1.

Teacher's Note:
1. All triple products are cyclic equal.

Answer: (d) 16

Explanation:
1. \( |\vec{a}\times\vec{b}|^2 = 16\cdot4\cdot\sin^2\frac\pi6 = 16 \).

Teacher's Note:
1. Use
2. a
3.
4. b
5. sin θ.

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

Explanation:
1. \( \vec{b}\times(\vec{a}\times\vec{b}) = \vec{a} - (\vec{a}\cdot\vec{b})\vec{b} \), so the sum is \( \vec{a} \).

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

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

Explanation:
1. \( \vec{b}\cdot\vec{a}\times\vec{c} = [\vec{b}\,\vec{a}\,\vec{c}] = -[\vec{a}\,\vec{b}\,\vec{c}] \); the others are cyclic or equal.

Teacher's Note:
1. Swapping two vectors changes the sign.

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

Explanation:
1. These are the components of \( \vec{a} \) along the axes, so the sum is \( \vec{a} \).

Teacher's Note:
1. It resolves a into components.

Answer: (a) \( (2\vec{a} + \vec{b}) \perp \vec{b} \)

Explanation:
1. \( (\vec{a} + \vec{b})\cdot\vec{a} = 0 \) gives \( \vec{a}\cdot\vec{b} = -|\vec{a}|^2 \).
2. \( (2\vec{a} + \vec{b})\cdot\vec{b} = -2|\vec{a}|^2 + 2|\vec{a}|^2 = 0 \).

Teacher's Note:
1. Use the given perpendicularity.

Answer: (c) \( \alpha = \pm1, \beta = 1 \)

Explanation:
1. Dependence: the determinant \( \begin{vmatrix}1&1&1\\4&3&4\\1&\alpha&\beta\end{vmatrix} = 1 - \beta = 0 \), so \( \beta = 1 \).
2. \( |\vec{c}|^2 = 2 + \alpha^2 = 3 \) gives \( \alpha = \pm1 \).

Teacher's Note:
1. Set the determinant to zero.
2. Then use the magnitude.

Answer: (d) 3

Explanation:
1. \( \vec{p}\cdot\vec{a} = \frac{[\vec{a}\,\vec{b}\,\vec{c}]}{[\vec{a}\,\vec{b}\,\vec{c}]} = 1 \) and similarly for the others, so the sum is 3.

Teacher's Note:
1. Each term is 1.

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

Explanation:
1. \( |\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2(\sin^2\theta + \cos^2\theta) \).

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

Answer: (c) \( \frac\pi6 \)

Explanation:
1. \( \vec{a}\times(\vec{a}\times\vec{c}) = (\vec{a}\cdot\vec{c})\vec{a} - \vec{c} \), so \( \vec{b} = \vec{c} - (\vec{a}\cdot\vec{c})\vec{a} \).
2. \( |\vec{b}|^2 = 4 - (\vec{a}\cdot\vec{c})^2 = 1 \), so \( \vec{a}\cdot\vec{c} = \sqrt3 = 2\cos\theta \) and \( \theta = \frac\pi6 \).

Teacher's Note:
1. Take the magnitude of b.

Answer: (c) the null vector

Explanation:
1. Let \( \vec{a} + \vec{b} = \lambda\vec{c} \) and \( \vec{b} + \vec{c} = \mu\vec{a} \). Subtracting, \( (1 + \mu)\vec{a} = (1 + \lambda)\vec{c} \).
2. As \( \vec{a}, \vec{c} \) are non-collinear, \( \mu = \lambda = -1 \), so \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \).

Teacher's Note:
1. Subtract the two collinearity relations.
2. Non-collinear means both coefficients vanish.

Answer: (b) \( \alpha = \frac\pi2, \beta = \frac\pi3 \)

Explanation:
1. \( (\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c} = \frac12\vec{b} \) gives \( \vec{a}\cdot\vec{c} = \frac12 \) and \( \vec{a}\cdot\vec{b} = 0 \).
2. So \( \alpha = \frac\pi2 \) and \( \beta = \frac\pi3 \).

Teacher's Note:
1. Compare coefficients.

Answer: (c) \( \frac1{\sqrt3} \)

Explanation:
1. The plane is \( x + y + z = 4 \). Distance \( = \frac{|3 - 4|}{\sqrt3} = \frac1{\sqrt3} \).

Teacher's Note:
1. Find the plane equation first.

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

Explanation:
1. The resolved part along \( \vec{b} \) is the projection \( \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\vec{b} \).

Teacher's Note:
1. Projection formula.

Answer: (a) \( \frac1{\sqrt{51}}(\hat{i} - 5\hat{j} + 5\hat{k}) \)

Explanation:
1. For \( \vec{x} = (p,q,r) \) equal dot products give \( 5p = q + 2r \) and \( 5p = -4q - 3r \), so \( q = -r \), \( r = 5p \).
2. \( \vec{x} \propto (1,-5,5) \), normalised by \( \sqrt{51} \).

Teacher's Note:
1. Equate the three dot products.

Answer: (a) \( \frac{\sqrt3}2 \)

Explanation:
1. \( \vec{a} \) is perpendicular to both \( \vec{b}, \vec{c} \), so \( |[\vec{a}\,\vec{b}\,\vec{c}]| = |\vec{a}||\vec{b}\times\vec{c}| = \sin\frac\pi3 = \frac{\sqrt3}2 \).

Teacher's Note:
1. a is parallel to b×c.

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

Explanation:
1. A cross product is perpendicular to both of its factors, in particular to \( \vec{a}\times\vec{b} \).

Teacher's Note:
1. Cross product is perpendicular to its factors.

Answer: (d) equilateral

Explanation:
1. \( \vec{p}\times\vec{q} \) is perpendicular to the plane of \( \vec{p}, \vec{q} \), so it is independent of \( \vec{p} \) and \( \vec{q} \).
2. Hence \( b - c = c - a = a - b = 0 \) and the triangle is equilateral.

Teacher's Note:
1. Three independent directions.

Answer: (b) \( \frac1x + \frac1y + \frac1z = 1 \)

Explanation:
1. Shift the origin to \( (2,2,2) \): the points are \( (-x,0,0), (0,-y,0), (0,0,-z) \) and \( D = (-1,-1,-1) \).
2. The plane \( \frac{u}{-x} + \frac{v}{-y} + \frac{w}{-z} = 1 \) contains D, so \( \frac1x + \frac1y + \frac1z = 1 \).

Teacher's Note:
1. Use the intercept form of the plane.