JEE Mathematics Points Direction Cosines and Direction Rations MCQs Set 01

Answer: (c) -1

Explanation:
1. \( \cos2\theta = 2\cos^2\theta - 1 \), so the sum is \( 2(\cos^2\alpha + \cos^2\beta + \cos^2\gamma) - 3 \).
2. Since \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \), the value is \( 2 - 3 = -1 \).

Teacher's Note:
1. The squares of direction cosines add to 1.
2. Use the double-angle formula for cosine.

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

Explanation:
1. \( \cos^2\gamma = 1 - \cos^2 60^\circ - \cos^2 45^\circ = 1 - \frac14 - \frac12 = \frac14 \).
2. \( \cos\gamma = \frac12 \) for the acute angle, so \( \gamma = 60^\circ \).

Teacher's Note:
1. Use l² + m² + n² = 1.
2. Choose the acute angle as asked.

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

Explanation:
1. The numerator is 1 and the denominator is \( 3 - 1 = 2 \).
2. \( \cos\theta = \frac12 \), so \( \theta = \frac\pi3 \).

Teacher's Note:
1. Convert sin² to 1 - cos².
2. The sum of squares of direction cosines is 1.

Answer: (a) \( \left(\frac{1}{\sqrt3}, -\frac12, \frac{\sqrt5}{2\sqrt3}\right) \)

Explanation:
1. \( n^2 = 1 - \frac13 - \frac14 = \frac{5}{12} \), so \( n = \frac{\sqrt5}{2\sqrt3} \).
2. At unit distance the coordinates equal the direction cosines.

Teacher's Note:
1. At unit distance, the point's coordinates are the direction cosines.
2. Take the positive root as n > 0.

Answer: (a) \( -\frac73 \)

Explanation:
1. Perpendicular: \( 1\cdot\lambda - 3(1+\lambda) + 5(2+\lambda) = 0 \).
2. \( 3\lambda + 7 = 0 \), so \( \lambda = -\frac73 \).

Teacher's Note:
1. Perpendicular lines have zero dot product of direction ratios.
2. Expand carefully.

Answer: (b) \( \frac{1}{\sqrt{30}}, -\sqrt{\frac56}, \sqrt{\frac{2}{15}} \)

Explanation:
1. Direction ratios: \( (3-2, -2-3, 1-(-1)) = (1, -5, 2) \), with magnitude \( \sqrt{30} \).
2. Direction cosines: \( \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}}, \frac{2}{\sqrt{30}} \), and \( \frac{5}{\sqrt{30}} = \sqrt{\frac56} \), \( \frac{2}{\sqrt{30}} = \sqrt{\frac{2}{15}} \).

Teacher's Note:
1. Divide the direction ratios by their magnitude.
2. Direction cosines must satisfy l² + m² + n² = 1.

Answer: (d) -2

Explanation:
1. \( \vec{OA} \cdot \vec{OB} = 0 \): \( -2 + 8 + 3\lambda = 0 \).
2. \( \lambda = -2 \).

Teacher's Note:
1. A right angle gives a zero dot product.
2. Use position vectors from the origin.

Answer: (c) \( \left(0, -\frac94, 0\right) \)

Explanation:
1. Let \( C = (0, y, 0) \). \( \vec{AB} = (-1, -4, 3) \) and \( \vec{BC} = (-1, y+1, -2) \).
2. \( AB \perp BC \): \( 1 - 4(y+1) - 6 = 0 \), so \( y = -\frac94 \).

Teacher's Note:
1. A point on the y-axis has x = z = 0.
2. Perpendicular means dot product zero.

Answer: (a) 0

Explanation:
1. \( \vec{AB} = (1, 4, 3) \) and \( \vec{AC} = (\lambda - 1, -4, -3) \).
2. Parallel: \( \frac{\lambda-1}{1} = \frac{-4}{4} = \frac{-3}{3} = -1 \), so \( \lambda = 0 \).

Teacher's Note:
1. Collinear points give proportional direction ratios.
2. Check all three components.

Answer: (b) 0

Explanation:
1. \( \vec{AB} = (1, -2, 2) \), \( \vec{AC} = (-1, 3, -3) \), \( \vec{AD} = (0, 1, \lambda - 1) \).
2. Coplanar: \( \begin{vmatrix} 1 & -2 & 2 \\ -1 & 3 & -3 \\ 0 & 1 & \lambda - 1 \end{vmatrix} = 0 \), which simplifies to \( \lambda = 0 \).

Teacher's Note:
1. Four points are coplanar if the scalar triple product of three edge vectors is zero.
2. Expand the determinant along the first column.

Answer: (c) \( \frac{1}{\sqrt3}\sum L, \frac{1}{\sqrt3}\sum M, \frac{1}{\sqrt3}\sum N \)

Explanation:
1. Treat the three lines as new axes. The line equally inclined to them has direction ratios \( (1, 1, 1) \) in that frame, i.e. unit vector \( \frac{1}{\sqrt3}(\hat u_1 + \hat u_2 + \hat u_3) \).
2. Its direction cosines are \( \frac{1}{\sqrt3}\sum L_r \), \( \frac{1}{\sqrt3}\sum M_r \), \( \frac{1}{\sqrt3}\sum N_r \) (the sum vector has magnitude \( \sqrt3 \)).

Teacher's Note:
1. The sum of three mutually perpendicular unit vectors has length √3.
2. Divide by √3 to normalise.

Answer: (d) \( \frac{1}{\sqrt3}, -\frac{1}{\sqrt3}, -\frac{1}{\sqrt3} \)

Explanation:
1. Unit vectors along the lines: \( \left(\frac13, \frac23, \frac23\right) \), \( \left(\frac27, \frac37, \frac67\right) \), \( \left(\frac{3}{13}, \frac{4}{13}, \frac{12}{13}\right) \).
2. For \( d = \frac{1}{\sqrt3}(1, -1, -1) \) the dot products are \( \frac{1-2-2}{3\sqrt3} \), \( \frac{2-3-6}{7\sqrt3} \), \( \frac{3-4-12}{13\sqrt3} \), all equal to \( -\frac{1}{\sqrt3} \).
3. The other options give unequal dot products (for option (c): \( \frac{5}{3\sqrt3}, \frac{11}{7\sqrt3}, \frac{19}{13\sqrt3} \)).

Teacher's Note:
1. Equal inclination means equal dot products with the unit vectors.
2. Test the options one by one.

Answer: (c) \( \frac\pi3 \)

Explanation:
1. Put \( n = -(l + m) \): \( l^2 = m^2 + (l+m)^2 \) gives \( m(m + l) = 0 \).
2. So \( m = 0, l = -n \): direction \( (1, 0, -1) \); or \( m = -l, n = 0 \): direction \( (1, -1, 0) \).
3. \( \cos\theta = \frac{1}{\sqrt2\cdot\sqrt2} = \frac12 \), so \( \theta = \frac\pi3 \).

Teacher's Note:
1. Eliminate one variable.
2. Two lines result; use the dot product for the angle.

Answer: (a) 0

Explanation:
1. \( \vec{BA} = (5, 10, 15) \) and \( \vec{BC} = (5, -10, \lambda + 5) \).
2. \( 25 - 100 + 15(\lambda + 5) = 0 \) gives \( 15\lambda = 0 \), so \( \lambda = 0 \).

Teacher's Note:
1. A right angle at B means BA · BC = 0.
2. Subtract B from the other points.

Answer: (c) parallelogram

Explanation:
1. \( \vec{AB} = (-2, -4, -4) \) and \( \vec{DC} = (-2, -4, -4) \), so \( AB \parallel DC \) and equal in length.
2. \( \vec{BC} = (3, 5, 3) \) and \( \vec{AB} \cdot \vec{BC} = -6 - 20 - 12 \neq 0 \), so the angle is not a right angle.
3. Hence ABCD is a parallelogram.

Teacher's Note:
1. Compare opposite sides as vectors.
2. Check one angle for a rectangle.

Answer: (c) rhombus

Explanation:
1. \( \vec{AB} = (2, -3, 6) = \vec{DC} \), so ABCD is a parallelogram.
2. \( |AB| = 7 \) and \( |BC| = |(-6, -2, 3)| = 7 \): adjacent sides are equal.
3. \( \vec{AB} \cdot \vec{BC} = -12 + 6 + 18 = 12 \neq 0 \), so it is not a square. It is a rhombus.

Teacher's Note:
1. Equal adjacent sides with no right angle give a rhombus.
2. Opposite sides must be equal vectors for a parallelogram.

Answer: (b) (2, 2, -1)

Explanation:
1. Centroid \( = \left(\frac{-1+2+5}{3}, \frac{6+1-1}{3}, \frac{-4+1+0}{3}\right) = (2, 2, -1) \).

Teacher's Note:
1. The centroid is the mean of the vertex coordinates.
2. Average each coordinate separately.

Answer: (a) (5, 1, 2)

Explanation:
1. \( C = 3G - A - B = (6 + 1 - 2,\ 0 - 2 + 3,\ 6 - 4 - 0) = (5, 1, 2) \).

Teacher's Note:
1. Sum of the three vertices equals three times the centroid.
2. Subtract the known vertices.

Answer: (c) (1, -2, 3)

Explanation:
1. Centroid \( = \left(\frac{0+4+0+0}{4}, \frac{0+0-8+0}{4}, \frac{0+0+0+12}{4}\right) = (1, -2, 3) \).

Teacher's Note:
1. For four points divide the sum by 4.
2. Do not divide by 3 for a tetrahedron.

Answer: (a) (2, -4, 18)

Explanation:
1. Sum of the four vertices \( = 4(1, -2, 5) = (4, -8, 20) \).
2. Fourth vertex \( = (4 - 2, -8 + 4, 20 - 2) = (2, -4, 18) \).

Teacher's Note:
1. Use sum of vertices = 4 × centroid.
2. Subtract the three known points.

Answer: (b) vertices of a rectangle

Explanation:
1. \( \vec{AB} = (1, 3, -1) = \vec{DC} \) and \( \vec{AD} = (2, -1, -1) = \vec{BC} \), so ABCD is a parallelogram.
2. \( \vec{AB} \cdot \vec{AD} = 2 - 3 + 1 = 0 \), so adjacent sides are perpendicular; \( |AB| = \sqrt{11} \neq |AD| = \sqrt6 \).
3. It is a rectangle but not a square or rhombus.

Teacher's Note:
1. Test the parallelogram, then the right angle, then the equal sides.
2. Unequal sides rule out square and rhombus.

Answer: (d) 13

Explanation:
1. Length \( = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{169} = 13 \).

Teacher's Note:
1. Projections on axes are the coordinate differences.
2. Use the 3-D distance formula.

Answer: (b) \( \lambda = 7, \mu = 10 \)

Explanation:
1. The midpoint of BC is \( M\left(\frac{\lambda - 1}{2}, 4, \frac{\mu + 2}{2}\right) \), so \( \vec{AM} = \left(\frac{\lambda - 5}{2}, 1, \frac{\mu - 8}{2}\right) \).
2. Equally inclined with the axes means equal direction ratios: \( \frac{\lambda-5}{2} = 1 = \frac{\mu - 8}{2} \), so \( \lambda = 7 \) and \( \mu = 10 \).

Teacher's Note:
1. A median goes to the midpoint of the opposite side.
2. Equal inclination means equal components.

Answer: (c) 6

Explanation:
1. Edges from \( (0, 1, 2) \): \( (4, 2, 4) \), \( (2, 2, 0) \), \( (3, -1, -1) \).
2. The determinant is \( 4(-2) - 2(-2) + 4(-2 - 6) = -36 \), so the volume is \( \frac{36}{6} = 6 \).

Teacher's Note:
1. Volume =
2. scalar triple product
3. / 6.
4. Take edges from a common vertex.

Answer: (a) (2, 0, -1)

Explanation:
1. Diagonals bisect each other, so \( A + C = B + D \).
2. \( D = A + C - B = (1 - 1 + 2,\ -1 + 2 - 1,\ 0 + 7 - 8) = (2, 0, -1) \).

Teacher's Note:
1. Diagonals of a parallelogram bisect each other.
2. D = A + C - B for consecutive vertices A, B, C.

Answer: (b) collinear points

Explanation:
1. Write \( P_r = (a\rho^{r-1}, b\rho^{r-1}, c\rho^{r-1}) \), so \( P_r = \rho^{r-1}(a, b, c) \).
2. All three points lie on the line through the origin with direction \( (a, b, c) \), so they are collinear.

Teacher's Note:
1. Each point is a scalar multiple of one fixed point.
2. Points on a line through the origin are collinear.

Answer: (a) (i) and (iii)

Explanation:
1. \( \vec{PA} = (-1, -2, 2) \) has length 3, so the unit direction is \( \pm\frac13(1, 2, -2) \).
2. \( Q = P \pm 3 \cdot \frac13(1, 2, -2) = (1, 1, 1) \pm (1, 2, -2) \), i.e. \( (2, 3, -1) \) or \( (0, -1, 3) \).

Teacher's Note:
1. Find the unit vector along the line, then move 3 units either way.
2. There are two possible points.

Answer: (b) (ii) and (iv)

Explanation:
1. Internal division in the ratio 3 : 2: \( P = \frac{2A + 3B}{5} = \left(\frac15, \frac65, -\frac15\right) \).
2. External division: \( P = \frac{3B - 2A}{3 - 2} = (-7, 18, -29) \).

Teacher's Note:
1. A ratio on a line can be internal or external.
2. External division uses subtraction in the section formula.

Answer: (d) (ii) and (iv)

Explanation:
1. \( \cos 60^\circ = \pm\frac{1 - \lambda}{\sqrt{(1+\lambda)^2 + (1-\lambda)^2 + 4}} = \pm\frac12 \), so \( 4(1-\lambda)^2 = 2\lambda^2 + 6 \).
2. This gives \( \lambda^2 - 4\lambda - 1 = 0 \), so \( \lambda = 2 \pm \sqrt5 \), which are options (ii) and (iv).

Teacher's Note:
1. A line has two opposite directions, so the cosine may be ±1/2.
2. Square both sides to remove the root.