JEE Mathematics Straight Lines in Space MCQs Set 02

Answer: (c) \( \frac{4}{3} \)

Explanation:
1. Let the line have direction cosines \( \ell, m, n \). The diagonals have direction cosines \( \frac{1}{\sqrt{3}}(\pm 1, \pm 1, \pm 1) \) with one sign fixed, giving four directions.
2. \( \cos^2\alpha = \frac{(\ell + m + n)^2}{3} \), and similarly with sign changes on one of \( \ell, m, n \) in turn.
3. Adding the four squares, the cross terms cancel: \( \frac{4(\ell^2 + m^2 + n^2)}{3} = \frac{4}{3} \).

Teacher's Note:
1. Write each cosine as a dot product of direction cosines.
2. The cross terms cancel in the sum of the four squares.

Answer: (c) passes through the origin and is perpendicular to the given lines is \( x = y = -z \)

Explanation:
1. Direction ratios \( (a, b, c) \) perpendicular to both lines satisfy \( 2a + 3b + 5c = 0 \) and \( a + 2b + 3c = 0 \).
2. By cross multiplication \( \frac{a}{9 - 10} = \frac{b}{5 - 6} = \frac{c}{4 - 3} \), i.e. \( (a, b, c) \propto (1, 1, -1) \).
3. So the line through the origin perpendicular to both is \( x = y = -z \). The angle bisector would have direction \( \hat{a} + \hat{b} \), which is not \( (3, 3, 8) \) or \( (1, 2, 3) \), so (a) and (b) are false.

Teacher's Note:
1. Cross multiplication gives the direction ratios of a line perpendicular to two given directions.
2. Check the bisector using the sum of the unit direction vectors.

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

Explanation:
1. \( L_1 \) has direction cosines \( \left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right) \).
2. Rotating the xz-plane by \( 90^\circ \) about the z-axis turns it into the yz-plane, so \( L_2 \) has direction cosines \( \left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \).
3. \( \cos\theta = 0 + 0 + \frac{1}{2} = \frac{1}{2} \), so \( \theta = 60^\circ \).

Teacher's Note:
1. The rotation about the z-axis keeps the z-component unchanged.
2. Use \( \cos\theta = \ell_1\ell_2 + m_1m_2 + n_1n_2 \).

Answer: (a) \( -2 \)

Explanation:
1. Four planes \( P_1: 2x - y + 3z + 4 = 0 \), \( P_2: ax + y - z + 2 = 0 \), \( P_3: x - 3y + z = 0 \), \( P_4: x + 2y + z + 1 = 0 \) define the two lines \( P_1 \cap P_2 \) and \( P_3 \cap P_4 \).
2. Two lines given as intersections of pairs of planes are coplanar exactly when the \( 4 \times 4 \) determinant of the coefficients (with constants) vanishes: \( \begin{vmatrix} 2 & -1 & 3 & 4 \\ a & 1 & -1 & 2 \\ 1 & -3 & 1 & 0 \\ 1 & 2 & 1 & 1 \end{vmatrix} = 0 \).
3. Expanding the determinant gives \( 12(a + 2) = 0 \), so \( a = -2 \).

Teacher's Note:
1. Use the four-plane determinant test when lines are given as intersecting planes.
2. The determinant factorises into a single linear factor in a.

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

Explanation:
1. \( \ell = \cos 45^\circ = \frac{1}{\sqrt{2}} \) and \( m = \cos 120^\circ = -\frac{1}{2} \).
2. \( \ell^2 + m^2 + n^2 = 1 \) gives \( n^2 = 1 - \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \).
3. For an acute angle \( n = \frac{1}{2} \), so \( \theta = 60^\circ \).

Teacher's Note:
1. Use \( \ell^2 + m^2 + n^2 = 1 \).
2. Acute angle means a positive cosine.

Answer: (a) \( \frac{2}{3} \)

Explanation:
1. The angle \( \theta \) between a line and a plane satisfies \( \sin\theta = \frac{|\vec{d}\cdot\vec{n}|}{|\vec{d}||\vec{n}|} = \frac{5 + 3\lambda}{\sqrt{14}\sqrt{5 + \lambda^2}} \).
2. \( \theta = \cos^{-1}\sqrt{\frac{5}{14}} \) gives \( \sin\theta = \frac{3}{\sqrt{14}} \).
3. \( 9(5 + \lambda^2) = (5 + 3\lambda)^2 \) gives \( 45 = 25 + 30\lambda \), so \( \lambda = \frac{2}{3} \).

Teacher's Note:
1. The sine, not the cosine, of the line-plane angle uses the dot product of direction and normal.
2. Convert the given cosine into a sine.

Answer: (c) \( \sqrt{53} \)

Explanation:
1. Let the foot be \( (2t, 3t + 2, 4t + 3) \). The vector from the given point is \( (2t - 3, 3t + 3, 4t - 8) \).
2. Perpendicular to direction \( (2, 3, 4) \): \( 2(2t - 3) + 3(3t + 3) + 4(4t - 8) = 0 \), so \( 29t = 29 \), \( t = 1 \).
3. Foot \( (2, 5, 7) \); distance \( = \sqrt{1 + 36 + 16} = \sqrt{53} \).

Teacher's Note:
1. Parametrise the foot of the perpendicular.
2. The connecting vector is perpendicular to the line's direction.

Answer: (c) \( \frac{9}{2} \)

Explanation:
1. Intersecting lines are coplanar: \( \begin{vmatrix} 3 - 1 & k + 1 & 0 - 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0 \).
2. Expanding: \( 2(3 - 8) - (k + 1)(2 - 4) - 1(4 - 3) = 0 \), i.e. \( -10 + 2(k + 1) - 1 = 0 \).
3. So \( 2(k + 1) = 11 \) and \( k = \frac{9}{2} \).

Teacher's Note:
1. Use the vector joining a point on each line.
2. Coplanarity of the connecting vector and both directions is necessary; for non-parallel lines it is also sufficient for intersection.

Answer: (c) exactly two values

Explanation:
1. Coplanarity: \( \begin{vmatrix} 1 & -1 & -1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{vmatrix} = 0 \), where the first row is the vector between the points.
2. Expanding along the first row: \( 1(1 + 2k) + 1(1 + k^2) - 1(2 - k) = 0 \).
3. This simplifies to \( k^2 + 3k = 0 \), so \( k = 0 \) or \( k = -3 \): exactly two values.

Teacher's Note:
1. Write the coplanarity determinant carefully.
2. Solve the resulting quadratic.

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

Explanation:
1. Substituting \( n = -\ell - m \) in \( \ell^2 = m^2 + n^2 \): \( \ell^2 = m^2 + (\ell + m)^2 \), so \( 2m(m + \ell) = 0 \).
2. Case \( m = 0 \): \( n = -\ell \), direction ratios \( (1, 0, -1) \). Case \( \ell = -m \): \( n = 0 \), direction ratios \( (-1, 1, 0) \).
3. \( \cos\theta = \frac{|-1 + 0 + 0|}{\sqrt{2}\sqrt{2}} = \frac{1}{2} \), so \( \theta = \frac{\pi}{3} \).

Teacher's Note:
1. Eliminate one direction cosine to get two directions.
2. Use the dot product of the two direction ratio sets.

Answer: (d) 13

Explanation:
1. A point on the line is \( (3\lambda + 2, 4\lambda - 1, 12\lambda + 2) \).
2. Substitute in the plane: \( 3\lambda + 2 - 4\lambda + 1 + 12\lambda + 2 = 16 \), so \( 11\lambda = 11 \) and \( \lambda = 1 \): the point is \( (5, 3, 14) \).
3. Distance from \( (1, 0, 2) \): \( \sqrt{16 + 9 + 144} = 13 \).

Teacher's Note:
1. Parametrise the line and substitute into the plane.
2. Then apply the distance formula.

Answer: (c) 2

Explanation:
1. The point \( (3, -2, -4) \) lies on the plane: \( 3\ell - 2m + 4 = 9 \), so \( 3\ell - 2m = 5 \).
2. The direction is perpendicular to the normal \( (\ell, m, -1) \): \( 2\ell - m - 3 = 0 \).
3. Solving: \( \ell = 1,\ m = -1 \), so \( \ell^2 + m^2 = 2 \).

Teacher's Note:
1. A line lies in a plane if a point of it lies on the plane and its direction is perpendicular to the normal.
2. Two linear equations fix \( \ell \) and \( m \).

Answer: (d) \( aa' + c + c' = 0 \)

Explanation:
1. The first line is \( \frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c} \), with direction ratios \( (a, 1, c) \).
2. The second is \( \frac{x - b'}{a'} = \frac{y - d'}{c'} = \frac{z}{1} \), with direction ratios \( (a', c', 1) \).
3. Perpendicular lines: \( aa' + c' + c = 0 \).

Teacher's Note:
1. Write both lines in symmetric form to read off direction ratios.
2. Perpendicular lines have zero dot product of direction ratios.

Answer: (c) \( \frac{x + 3}{3} = \frac{y - 5}{1} = \frac{z - 2}{-5} \)

Explanation:
1. The direction \( (3, 1, -5) \) satisfies \( 2(3) - 1 + (-5) = 0 \), so the line is parallel to the plane and its image has the same direction.
2. The image of the point \( (1, 3, 4) \): \( \frac{x - 1}{2} = \frac{y - 3}{-1} = \frac{z - 4}{1} = \frac{-2(2 - 3 + 4 + 3)}{6} = -2 \), giving \( (-3, 5, 2) \).
3. The image line is \( \frac{x + 3}{3} = \frac{y - 5}{1} = \frac{z - 2}{-5} \).

Teacher's Note:
1. A line parallel to a plane has an image with the same direction.
2. Reflect one point of the line in the plane.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The midpoint of AB is \( M(1, 3, 5) \). For \( x = 1 \) the parameter is 1 and the line point is \( (1, 3, 5) \), so the line passes through M and R is true.
2. Direction ratios of AB are \( (0, 6, -4) \); with the line's \( (1, 2, 3) \): \( 0 + 12 - 12 = 0 \), so AB is perpendicular to the line. Hence A is the mirror image of B and A is true.
3. R only states that the line passes through the midpoint; the mirror image also needs perpendicularity, so R does not explain A.

Teacher's Note:
1. A mirror image needs both the midpoint on the line and a perpendicular join.
2. Passing through the midpoint alone is not enough.