CBSE Class 12 Mathematics Three Dimensional Geometry Previous Year Questions Set 01

Very Short Answer Questions (1 Mark)

Question 1. Write the direction cosines of a line equally inclined to the three coordinate axes. [CBSE 2009, 1M]
Answer:
We know that \( l^2 + m^2 + n^2 = 1 \)
given l = m = n
\( 3n^2 = 1 \Rightarrow n = \pm\frac{1}{\sqrt{3}} \)
d.c.'s are \( \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \) and \( \left(\frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right) \) [1]

Question 2. Find the direction cosines of a line which makes equal angles with the coordinate axes. [CBSE 2019, 1M]
OR
A line passes through the point with position vector \( 2\hat{i} - \hat{j} + 4\hat{k} \) and is in the direction of the vector \( \hat{i} + \hat{j} - 2\hat{k} \). Find the equation of the line in cartesian form.
Answer:
Let the line make angles α, β, γ with the OX, OY, OZ axes respectively.
Given α = β = γ
∵ \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \)
⇒ \( 3\cos^2\alpha = 1 \)   (∵ α = β = γ)
or \( \cos^2\alpha = \frac{1}{3} \)
⇒ \( \cos\alpha = \pm\frac{1}{\sqrt{3}} \)
∴ Direction cosines are \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \) or \( \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}} \) [1]
OR
Since the line passes through the point (2, −1, 4) and its direction ratios are 1, 1, −2,
so the equation of the line in cartesian form is:
\( \frac{x - 2}{1} = \frac{y + 1}{1} = \frac{z - 4}{-2} \) [1]

Question 3. The coordinates of the foot of the perpendicular drawn from the point (2, −3, 4) on the y-axis is [CBSE 2020, 1M]
(A) (2, 3, 4)    (B) (−2, −3, −4)    (C) (0, −3, 0)    (D) (2, 0, 4)
Answer:
Ans. (C)
The coordinates of the foot of the perpendicular on the y-axis from the point (2, −3, 4) are (0, −3, 0). [1]

Short Answer Questions (2 Marks)

Question 4. The x-coordinate of a point on the line joining the points P(2, 2, 1) and Q(5, 1, −2) is 4. Find its z-coordinate. [CBSE 2017, 2M]
Answer:

Let the required point be R(x, y, z).
∵ x-coordinate of R is 4
⇒ \( \frac{5\lambda + 2 \times 1}{\lambda + 1} = 4 \)
⇒ 5λ + 2 = 4λ + 4 ⇒ λ = 2 [1]
∴ z-coordinate of R
\( = \frac{2 \times (-2) + 1 \times 1}{2 + 1} = \frac{-3}{3} = -1 \) [1]

Long Answer Questions I (4 Marks)

Question 5. Find the points on the line \( \frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2} \) at a distance of 5 units from the point P(1, 3, 3). [CBSE 2010, 4M]
Answer:
\( \frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2} = \lambda \)
Any point on the line will be
(3λ − 2, 2λ − 1, 2λ + 3) [1]
Given that
\( \sqrt{(3\lambda - 2 - 1)^2 + (2\lambda - 1 - 3)^2 + (2\lambda + 3 - 3)^2} = 5 \) [1]
⇒ \( 9\lambda^2 - 18\lambda + 9 + 4\lambda^2 - 16\lambda + 16 + 4\lambda^2 = 25 \)
⇒ \( 17\lambda^2 - 34\lambda = 0 \) [1]
⇒ \( \lambda^2 - 2\lambda = 0 \) ⇒ λ = 0, 2
Hence the points are (−2, −1, 3) or (4, 3, 7). [1]

Question 6. Find the shortest distance between the following lines:
\( \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k}) \)
\( \vec{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(4\hat{i} + 6\hat{j} + 8\hat{k}) \) [CBSE 2015, 4M]
Answer:
\( \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k}) \)
\( \vec{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + 2\mu(2\hat{i} + 3\hat{j} + 4\hat{k}) \)
The lines are parallel.
Here \( \vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k} \)
\( \vec{a}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k} \)
\( \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) [1]
∴ \( \vec{a}_2 - \vec{a}_1 = \hat{i} + 2\hat{j} + 2\hat{k} \) and \( |\vec{b}| = \sqrt{4 + 9 + 16} = \sqrt{29} \) [1]
⇒ \( \vec{b} \times (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & 2 & 2 \end{vmatrix} \)
\( = \hat{i}(6 - 8) - \hat{j}(4 - 4) + \hat{k}(4 - 3) = -2\hat{i} + \hat{k} \) [1]
∴ \( |\vec{b} \times (\vec{a}_2 - \vec{a}_1)| = \sqrt{4 + 1} = \sqrt{5} \)
⇒ Required distance \( = \frac{|\vec{b} \times (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}|} = \frac{\sqrt{5}}{\sqrt{29}} \) [1]

Question 7. Find the coordinates of the foot of the perpendicular drawn from the point A(−1, 8, 4) to the line joining the points B(0, −1, 3) and C(2, −3, −1). Hence find the image of the point A in the line BC. [CBSE 2016, 4M]
Answer:
Equation of BC:
\( \frac{x}{2} = \frac{y + 1}{-2} = \frac{z - 3}{-4} = \lambda \) (let) [1]

Let the foot of the perpendicular be D(2λ, −2λ − 1, −4λ + 3).
Direction ratios of line AD are 2λ + 1, −2λ − 9, −4λ − 1
Direction ratios of line BC are 2, −2, −4 [1]
∵ AD ⊥ BC
∴ 2(2λ + 1) − 2(−2λ − 9) − 4(−4λ − 1) = 0
⇒ 24λ + 24 = 0
or λ = −1
∴ D(−2, 1, 7) [1]
Let A′(α, β, γ) be the image of A in the line BC, i.e. D is the mid-point of AA′.
∴ \( \frac{-1 + \alpha}{2} = -2; \ \frac{8 + \beta}{2} = 1; \ \frac{4 + \gamma}{2} = 7 \)
⇒ α = −3, β = −6, γ = 10
∴ A′(−3, −6, 10) [1]

Question 8. Find the shortest distance between the lines \( \vec{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k}) \) and \( \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} + 4\hat{j} - 5\hat{k}) \). [CBSE 2018, 4M]
Answer:
The equations of the lines are
\( \vec{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k}) \) and \( \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} + 4\hat{j} - 5\hat{k}) \) [1]
where \( \vec{a}_1 = 4\hat{i} - \hat{j}, \ \vec{b}_1 = \hat{i} + 2\hat{j} - 3\hat{k} \)
\( \vec{a}_2 = \hat{i} - \hat{j} + 2\hat{k}, \ \vec{b}_2 = 2\hat{i} + 4\hat{j} - 5\hat{k} \)
⇒ \( \vec{a}_2 - \vec{a}_1 = -3\hat{i} + 2\hat{k} \)
and \( \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} \) [1]
∴ \( \vec{b}_1 \times \vec{b}_2 = 2\hat{i} - \hat{j}, \ |\vec{b}_1 \times \vec{b}_2| = \sqrt{5} \)
⇒ Required S.D. \( = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| \)
\( = \left| \frac{(-3\hat{i} + 2\hat{k}) \cdot (2\hat{i} - \hat{j})}{\sqrt{5}} \right| = \left| \frac{-6}{\sqrt{5}} \right| = \frac{6}{\sqrt{5}} \) [2]

Question 9. If the lines \( \frac{x - 1}{-3} = \frac{y - 2}{2\lambda} = \frac{z - 3}{2} \) and \( \frac{x - 1}{3\lambda} = \frac{y - 1}{2} = \frac{z - 6}{-5} \) are perpendicular, find the value of λ. Hence, find whether the lines are intersecting or not. [CBSE 2019, 4M]
Answer:
The equations of the lines are:
\( \frac{x - 1}{-3} = \frac{y - 2}{2\lambda} = \frac{z - 3}{2} \) and \( \frac{x - 1}{3\lambda} = \frac{y - 1}{2} = \frac{z - 6}{-5} \)
If the given lines are perpendicular, then
(−3) × 3λ + (2λ) × 2 + 2 × (−5) = 0
⇒ (−9λ) + 4λ − 10 = 0
⇒ −5λ = 10
⇒ λ = −2 [2]
For λ = −2, the lines are \( \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + s(-3\hat{i} - 4\hat{j} + 2\hat{k}) \) and \( \vec{r} = (\hat{i} + \hat{j} + 6\hat{k}) + t(-6\hat{i} + 2\hat{j} - 5\hat{k}) \).
Here \( \vec{a}_2 - \vec{a}_1 = -\hat{j} + 3\hat{k} \) and \( \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -4 & 2 \\ -6 & 2 & -5 \end{vmatrix} = 16\hat{i} - 27\hat{j} - 30\hat{k} \)
Shortest distance \( = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \frac{|0 + 27 - 90|}{\sqrt{256 + 729 + 900}} = \frac{63}{\sqrt{1885}} \neq 0 \) [1½]
Hence, the lines are not intersecting. [½]

Question 10. Find the shortest distance between the lines
\( \vec{r} = 2\hat{i} - \hat{j} + \hat{k} + \lambda(3\hat{i} - 2\hat{j} + 5\hat{k}) \)
\( \vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \mu(4\hat{i} - \hat{j} + 3\hat{k}) \) [CBSE 2020, 4M]
Answer:
Given lines
\( \vec{r} = 2\hat{i} - \hat{j} + \hat{k} + \lambda(3\hat{i} - 2\hat{j} + 5\hat{k}) \)
\( \vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \mu(4\hat{i} - \hat{j} + 3\hat{k}) \)
Here \( \vec{a}_1 = 2\hat{i} - \hat{j} + \hat{k}, \ \vec{b}_1 = 3\hat{i} - 2\hat{j} + 5\hat{k} \)
\( \vec{a}_2 = 3\hat{i} + 2\hat{j} - 4\hat{k}, \ \vec{b}_2 = 4\hat{i} - \hat{j} + 3\hat{k} \)
⇒ \( (\vec{a}_2 - \vec{a}_1) = \hat{i} + 3\hat{j} - 5\hat{k} \)
and \( \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 5 \\ 4 & -1 & 3 \end{vmatrix} \)
\( \vec{b}_1 \times \vec{b}_2 = -\hat{i} + 11\hat{j} + 5\hat{k} \)
∴ \( |\vec{b}_1 \times \vec{b}_2| = \sqrt{1 + 121 + 25} = \sqrt{147} \) [1]
Shortest distance between the given lines is
\( D = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| \) [1]
\( = \left| \frac{(\hat{i} + 3\hat{j} - 5\hat{k}) \cdot (-\hat{i} + 11\hat{j} + 5\hat{k})}{\sqrt{147}} \right| \) [1]
\( = \left| \frac{-1 + 33 - 25}{\sqrt{147}} \right| = \frac{7}{\sqrt{147}} \) units [1]