CBSE Class 12 Mathematics Three Dimensional Geometry Class Test Set 01

Class Test Questions with Solutions

Question 1. 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. [1]
Answer:

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

Question 2. Vector form of the line \( \frac{x - 3}{-2} = \frac{y - 4}{-5} = \frac{z + 7}{13} \) is: [1]
Answer:
Cartesian equation of the line is
\( \frac{x - 3}{-2} = \frac{y - 4}{-5} = \frac{z + 7}{13} \)
Now, the vector equation will be
\( \vec{r} = (3\hat{i} + 4\hat{j} - 7\hat{k}) + \lambda(-2\hat{i} - 5\hat{j} + 13\hat{k}) \)

Question 3. If the cartesian equation of the line is \( \frac{3 - x}{5} = \frac{y + 4}{7} = \frac{2z - 6}{4} \), write the vector equation for the line. [1]
Answer:
Cartesian equation of the line is \( \frac{3 - x}{5} = \frac{y + 4}{7} = \frac{2z - 6}{4} \)
We can write it as \( \frac{x - 3}{-5} = \frac{y + 4}{7} = \frac{z - 3}{2} \)
So the vector equation is \( \vec{r} = (3\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda(-5\hat{i} + 7\hat{j} + 2\hat{k}) \), where λ is a constant.

Question 4. Find the equation of the line passing through the point (2, −1, 1) and parallel to the line \( \frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{-3} \). [2]
Answer:
Given that the line passes through the point (2, −1, 1) and is parallel to the line \( \frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{-3} \).
The direction ratios of the above line are 2, 7 and −3, so the direction ratios of the required line will be 2k, 7k and −3k.
Required line is \( \vec{r} = \vec{a} + \lambda\vec{b} \)
⇒ \( \vec{r} = 2\hat{i} - \hat{j} + \hat{k} + \lambda(2k\hat{i} + 7k\hat{j} - 3k\hat{k}) \)
or \( \vec{r} = 2\hat{i} - \hat{j} + \hat{k} + \lambda'(2\hat{i} + 7\hat{j} - 3\hat{k}) \)   [where λ′ = kλ]

Question 5. Find the length and the foot of the perpendicular drawn from the point (2, −1, 5) to the line \( \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11} \). [3]
Answer:
Given equation of the line is
\( \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11} = \lambda \) (say) ...(1)

Let N be the foot of the perpendicular from the point P(2, −1, 5) to the line (1).
∴ Coordinates of N are (10λ + 11, −4λ − 2, −11λ − 8)
Direction ratios of PN are
10λ + 11 − 2, −4λ − 2 + 1, −11λ − 8 − 5
= 10λ + 9, −4λ − 1, −11λ − 13
and direction ratios of line (1) are 10, −4, −11. Since PN ⊥ line (1),
⇒ 10(10λ + 9) − 4(−4λ − 1) − 11(−11λ − 13) = 0   \( [\because a_1a_2 + b_1b_2 + c_1c_2 = 0] \)
or 100λ + 90 + 16λ + 4 + 121λ + 143 = 0
⇒ 237λ + 237 = 0
or λ = −1
Hence the foot of the perpendicular is N(1, 2, 3)
and \( PN = \sqrt{(2 - 1)^2 + (-1 - 2)^2 + (5 - 3)^2} \)
\( = \sqrt{1 + 9 + 4} = \sqrt{14} \) units

Question 6. Find the vector equation of the line passing through the point (1, 2, −4) and perpendicular to the two lines \( \frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7} \) and \( \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5} \). [5]
Answer:
Let the required line be L, which passes through (1, 2, −4) and is ⊥ to the given two lines
\( \frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7} \) and \( \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5} \)
Let \( \vec{n}_1 = 3\hat{i} - 16\hat{j} + 7\hat{k} \) and \( \vec{n}_2 = 3\hat{i} + 8\hat{j} - 5\hat{k} \)
Now \( L \parallel (\vec{n}_1 \times \vec{n}_2) \)
\( L = \lambda'(\vec{n}_1 \times \vec{n}_2) \)
\( = \lambda' \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} \)
\( = \lambda'(24\hat{i} + 36\hat{j} + 72\hat{k}) \)
The equation of the required line is given as \( \vec{r} - \vec{a} = \lambda\vec{b} \)
⇒ \( \vec{r} - (\hat{i} + 2\hat{j} - 4\hat{k}) = \lambda'(24\hat{i} + 36\hat{j} + 72\hat{k}) \)
\( \vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \)
where λ = 12λ′