JEE Mathematics Three Dimensional Geometry MCQs Set 12

Answer: (b) Y-axis

Explanation:
1. The line lies in the plane y = c, and \( \frac{x-a}{l}=\frac{z-b}{m} \) gives d.r.'s (l, 0, m).
2. The y-component is 0, so the dot product with (0, 1, 0) is zero.
3. Hence the line is perpendicular to the Y-axis.

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (a) aa1 + cc1 + 1 = 0

Explanation:
1. Taking z as the parameter, the d.r.'s are (a, c, 1) and (a1, c1, 1).
2. Perpendicular lines have zero dot product: \( aa_1+cc_1+1=0 \).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (b) \( \frac{x-b}{a}=\frac y1=\frac{z-d}{c} \)

Explanation:
1. Take y as the parameter: \( \frac{x-b}{a}=y=\frac{z-d}{c} \).
2. This is the second option.

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (b) \( \frac{x-\alpha}{0}=\frac{y-\beta}{1}=\frac{z-\gamma}{0} \)

Explanation:
1. The Y-axis has d.r.'s (0, 1, 0).
2. So the line is \( \frac{x-\alpha}{0}=\frac{y-\beta}{1}=\frac{z-\gamma}{0} \).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (b) al + bm + cn = 0

Explanation:
1. The plane passes through (x1, y1, z1).
2. The line lies in it, so its direction is perpendicular to the normal: \( al+bm+cn=0 \).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (b) \( \frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1} \)

Explanation:
1. The z-axis has d.r.'s (0, 0, 1).
2. The line is \( \frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1} \).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

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

Explanation:
1. Equal inclinations mean equal direction cosines, so the d.r.'s are (1, 1, 1).
2. The line is \( \frac{x-3}{1}=\frac{y-1}{1}=\frac{z-2}{1} \).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (c) \( \frac{x-3}{3}=\frac{y}{1}=\frac{z-5}{9} \)

Explanation:
1. A parallel line has the same d.r.'s (3, 1, 9).
2. Through (3, 0, 5): \( \frac{x-3}{3}=\frac y1=\frac{z-5}{9} \).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

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

Explanation:
1. A line perpendicular to a plane is along its normal (2, 3, 1).
2. The line is \( \frac{x+1}{2}=\frac{y-2}{3}=\frac{z+3}{1} \).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (b) \( \vec r=(5\hat i-4\hat j+6\hat k)+\lambda(3\hat i+7\hat j+2\hat k) \)

Explanation:
1. The point on the line is (5, -4, 6) and the d.r.'s are (3, 7, 2).
2. So \( \vec r=(5\hat i-4\hat j+6\hat k)+\lambda(3\hat i+7\hat j+2\hat k) \).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (a) \( \frac{70}{11} \)

Explanation:
1. Rewrite in standard form: the first has d.r.'s \( \left(-3,\frac{2p}{7},2\right) \) and the second \( \left(-\frac{3p}{7},1,-5\right) \).
2. Perpendicular: \( \frac{9p}{7}+\frac{2p}{7}-10=0 \).
3. \( \frac{11p}{7}=10 \), so \( p=\frac{70}{11} \).

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (c) \( 45^\circ \)

Explanation:
1. \( \cos\theta=\frac{2+1+0}{3\sqrt2}=\frac1{\sqrt2} \).
2. So the angle is 45 degrees.

Teacher's Note:
1. A line is fixed by a point and a direction, so identify both first.
2. Use the dot product for perpendicularity and the cross product for common perpendiculars.

Answer: (c) \( \frac1{5\sqrt2} \)

Explanation:
1. \( \sin\theta=\frac{|3(2)+4(-2)+5(1)|}{\sqrt{50}\cdot3}=\frac{3}{15\sqrt2} \).
2. This equals \( \frac1{5\sqrt2} \).

Teacher's Note:
1. For a line and a plane the numerator is the dot product of the direction with the normal and the result is a sine, not a cosine.
2. Simplify the radical at the end.

Answer: (d) \( 90^\circ \)

Explanation:
1. Divide by 6: \( \frac x3=\frac y2=\frac z{-6} \) and (dividing by 12) \( \frac x2=\frac y{-12}=\frac z{-3} \).
2. Dot product: 6 - 24 + 18 = 0.
3. So the lines are perpendicular.

Teacher's Note:
1. Convert to symmetric form with 1 as the coefficient of each variable.
2. Check for a zero dot product first.

Answer: (b) a = 6, b = 4

Explanation:
1. The line is \( (5-2t,\ 1+t(b-1),\ a+t(1-a)) \).
2. x = 0 gives t = 5/2.
3. \( 1+\frac52(b-1)=\frac{17}2 \) gives b = 4, and \( a+\frac52(1-a)=-\frac{13}2 \) gives a = 6.

Teacher's Note:
1. Write the line in parametric form so that every point on it is a function of one parameter t.
2. Then impose the given condition to find t.

Answer: (a) \( \frac\pi2 \)

Explanation:
1. The line x = 1, y = 2 is parallel to the z-axis.
2. The line y = -1, z = 0 is parallel to the x-axis.
3. The axes are perpendicular, so the angle is \( \frac\pi2 \).

Teacher's Note:
1. Two equations fixing two coordinates give a line parallel to the third axis.
2. Compare the axes directly.

Answer: (b) 2

Explanation:
1. The line joining the given points is (1, 1, 1).
2. Coplanarity: \( \begin{vmatrix}1&1&1\\a&3&4\\3&4&5\end{vmatrix}=0 \).
3. Expanding: \( (15-16)-(5a-12)+(4a-9)=2-a=0 \), so a = 2.

Teacher's Note:
1. Coplanar lines have a zero scalar triple product of the join vector and the two directions.
2. Expand the determinant along the first row.

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

Explanation:
1. A point on the line: \( (3t-3,\ 2-2t,\ t-1) \).
2. Substituting in the plane: \( 12t-12+10-10t+3t-3-5=0 \), so 5t = 10, t = 2.
3. The point is (3, -2, 1).

Teacher's Note:
1. Write the line in parametric form so that every point on it is a function of one parameter t.
2. Then impose the given condition to find t.

Answer: (c) exactly two values

Explanation:
1. Join vector between (2, 3, 4) and (1, 4, 5): (-1, 1, 1).
2. \( \begin{vmatrix}-1&1&1\\1&1&-k\\k&2&1\end{vmatrix}=-(k^2+3k)=0 \).
3. So k = 0 or k = -3.

Teacher's Note:
1. Set the scalar triple product to zero.
2. A quadratic gives up to two values.

Answer: (b) (3a, 2a, 3a); (a, a, a)

Explanation:
1. Points: P = (t, t - a, t) and Q = (s - a, s/2, s/2).
2. PQ is proportional to (2, 1, 2): from the y and z components, a = -k and, with the x component, s - t = -a and s/2 - t = -2a.
3. So s = 2a and t = 3a, giving P = (3a, 2a, 3a) and Q = (a, a, a).

Teacher's Note:
1. Parametrise both lines and make PQ proportional to the given direction.
2. Solve the resulting equations for the two parameters.

Answer: (c) \( \frac92 \)

Explanation:
1. Join vector (3,k,0) - (1,-1,1) = (2, k + 1, -1).
2. \( \begin{vmatrix}2&k+1&-1\\2&3&4\\1&2&1\end{vmatrix}=2(3-8)-(k+1)(2-4)-1(4-3)=2k-9=0 \).
3. So \( k=\frac92 \).

Teacher's Note:
1. Intersecting lines are coplanar, so the same determinant applies.
2. Expand and solve the linear equation.

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

Explanation:
1. Direction of L: (2, 3, 1) x (1, 3, 2) = (3, -3, 3), i.e. (1, -1, 1).
2. \( \cos\alpha=\frac{1}{\sqrt{1+1+1}} \) (positive x-axis).

Teacher's Note:
1. The line of intersection is perpendicular to both normals.
2. Use the cross product for its direction.