Math Assignment Class XII Ch-11 | 3-Dimensional Geometry

Three Dimensional Geometry
Math. Assignment Class XII Chapter 11

Important questions on Three dimensional geometry chapter 11 class XII based on the board examination point of view with answers. Revision questions based on three dimensional geometry class XII

Extra Questions of Chapter 11 Three Dimensional Geometry

Question 1:

A line makes angle 𝜶, 𝛃, 𝛾 with x-axis, y-axis, z-axis. Then find cos2𝜶 + cos2𝛃 + cos2𝛾

Answer: -1
Question 2:

Write vector equation of the line: $\frac{2x-1}{\sqrt{3}} =\frac{y+2}{2} =\frac{z-3}{3}$
Question 3:

Find the Cartesian equation of the line which passes through the point (-2, 4, - 5) and is parallel to the line   $\frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}$
Hint: Two parallel lines have same direction ratio.
Answer: $\frac{x+2}{3}=\frac{y-4}{-5}= \frac{z+5}{6}$
Question 4: Find angle between pair of lines
$\frac{2-x}{-2}=\frac{y-1}{7}=\frac{z+3}{-3}$ and
$\frac{x+2}{-1}=\frac{2y-8}{4}=\frac{z-5}{4}$
Answer: 90^o
Question 5:

Find the point on the given line at a distance 3√2 from the point (1, 2, 3).
$\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$
Answer: $\lambda =0, \lambda =\frac{30}{17}$ and the point is either
$A(-2,-1,3)\; or\; A\left (\frac{56}{17},\frac{43}{17},\frac{111}{17} \right )$

Question 6: Find the shortest distance between the lines
$\vec{r}=(1+\lambda)\hat{i}+(2-\lambda)\hat{j}+(\lambda +1)\hat{k}$
$\vec{r}=(2\hat{i}-\hat{j}-\hat{k})+\mu (2\hat{i}+\hat{j}+2\hat{k})$
Answer:    $\vec{b_{1}}\times \vec{b_{2}}=-3\hat{i}+3\hat{k},\; Distance\; =\frac{3\sqrt{2}}{2}$
Question 7:
Find the equation of a line passing through the point P(2, -1, 3) and perpendicular to the lines:
$\vec{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda (2\hat{i}-2\hat{j}+\hat{k})$ and
$\vec{r}=(2\hat{i}-\hat{j}-3\hat{k})+\mu(\hat{i}+2\hat{j}+2\hat{k})$
Answer:   $\vec{r}=\left ( 2\hat{i}-\hat{j}+3\hat{k} \right )+\lambda \left ( 2\hat{i}+\hat{j}-2\hat{k} \right )$
Question 8: Show that the following lines are coplanar.
$\frac{5-x}{-4}=\frac{y-7}{4}=\frac{z+3}{-5}$

$\frac{x-8}{7}=\frac{2y-8}{2}=\frac{z-5}{3}$

Solution Hint
Two lines are coplanar if $(\overrightarrow{a}_{2}-\overrightarrow{a}_{1}).(\overrightarrow{b}_{1}\times\overrightarrow{b}_{2})=0$

Question 9
A line passing through (2, -1, 3) and is perpendicular to the lines
$\overrightarrow{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2\hat{i}-2\hat{j}+\hat{k})$ and
$\overrightarrow{r}=(\hat{2i}-\hat{j}-\hat{3k})+\mu(\hat{i}+2\hat{j}+2\hat{k})$
Obtain its equation in vector and Cartesian form.

Answer:
$\overrightarrow{r}=(\hat{2i}-\hat{j}+\hat{3k})+t(\hat{2i}+\hat{j}-2\hat{k})$ where t = 3λ

$\frac{x-2}{2}=\frac{y+1}{1}=\frac{z-3}{-2}=t$
Question 10
Find the distance between the lines given by

$\overrightarrow{r}=(\hat{i}+\hat{2j}-\hat{4k})+\lambda(\hat{2i}+\hat{3j}+\hat{6k})$ and

$\overrightarrow{r}=(\hat{3i}+\hat{3j}-\hat{5k})+\mu(\hat{4i}+\hat{6j}+\hat{12k})$

Answer: $\frac{\sqrt{293}}{7}$

Question 11

Find the coordinates of the image of the point (1, 6, 3) with respect to the line

$\overrightarrow{r}=(\hat{j}+2\hat{k})+\lambda(\hat{i}+2\hat{j}+3\hat{k})$ where λ is a scalar. Also find the distance of the image from the y-axis.

Answer:

Coordinates of foot of perpendicular are (1, 3, 5)

Coordinates of image of point P are (1, 0, 7)

Solution Hint

Algorithm to solve the question

Take any arbitrary point on the line AB in terms of λ and let it is the coordinate of point L
Find the direction ratios of line PL
Find direction ratio of line AB
Line AB 丄 PQ so use perpendicularity condition to find the value of λ
Putting the value of λ to find the coordinates of Point L.
Point L is the foot of perpendicular in line AB.
Let us take point Q the image of point P in the line AB.
So point L is the mid point of PQ, and by using mid point formula find the coordinates of point Q

Question 12

Find the length and foot of perpendicular drawn from the point (2, -1, 5) on the line $\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}$
Answer: $\lambda =-1, Q \left ( 1,2,3 \right ), \; \; \left | PQ \right |=\sqrt{14}$

Question 13

Find the coordinates of the foot of the perpendicular drawn from the point P(0, 2, 3) to the line

$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$

Answer: Coordinates of Perpendicular (2, 3, -1)

Question 14

Find the coordinate of the foot of perpendicular drawn from a point A(1, 8, 4) to the line joining the points B(0, -1, 3) and C(2, -3, -1)

Answer: Coordinate of foot of perpendicular (-5/3, 2/3, 19/3)

Question 15

Show that the lines

$\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and
$\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}$ intersect. Also find their point of intersection.

Ans: Yes these lines intersect and their point of intersection is (1/2, -1/2, -3/2)

Solution Algorithm:

Find arbitrary points on both the lines
Compairing x, y and z-coordinates we get three equations.
Solve the first two equations to find the values of λ and μ
Putting the values of λ and μ in equation three.
If we get LHS = RHS the the given lines intersect each other.
Putting the value of λ or μ in arbitrary points we get the point of intersection of two lines.

Question 16
Show that the following lines are intersecting, also find their point of intersection.
$\vec{r}=3\hat{i}+2\hat{j}-4\hat{k}+\lambda(\hat{i}+2\hat{j}+2\hat{k})$
$\vec{r}=5\hat{i}-2\hat{j}+\mu (3\hat{i}+2\hat{j}+6\hat{k})$
Answer: Point of intersection (-1, - 6, -12)
Solution Algorithm
Find   $(\overrightarrow{a}_{2}-\overrightarrow{a}_{1}).(\overrightarrow{b}_{1}\times\overrightarrow{b}_{2})$
If   $(\overrightarrow{a}_{2}-\overrightarrow{a}_{1}).(\overrightarrow{b}_{1}\times\overrightarrow{b}_{2})=0$ then lines are coplanar and $(\overrightarrow{b}_{1}\times\overrightarrow{b}_{2})\neq 0$ so lines are intersecting lines.
Question 17
Show that lines $\overrightarrow{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(\hat{3i}-\hat{j})$ and   $\overrightarrow{r}=(\hat{4i}-\hat{k})+\mu(\hat{2i}+\hat{3k})$ intersect. Also find their point of intersection.

Answer: Yes these lines intersect and the point of intersection is (4, 0, -1)

Question 18
An aeroplane is flying along the line   $\vec{r}=\lambda(\hat{i}-\hat{j}+\hat{k})$ where λ is a scalar and another aeroplane is flying along the line   $\vec{r}=\hat{i}-\hat{j}+\mu(-2\hat{j}+\hat{k})$   where μ is a scalar. At what points on the lines should they reach, so that the distance between them is the shortest ? Find the shortest possible distance between them.

Solution Hint:
Here we see that direction vectors of both the lines are different, so these lines are not parallel.
The given situation can be represented as in the figure given below.
Solution Algorithm
Find any point P(λ, -λ, λ) on AB
Find any point Q(1, -2μ - 1, μ) on CD
Find direction ratio's of PQ as (λ - 1, - λ + 2μ + 1, λ - μ)
PQ ⊥ CD so use perpendicularity condition to find eqn. (1) as 3λ - 3μ = 2
PQ ⊥ AB so use perpendicularity condition to find eqn. (2) as 3λ - 5μ = 2
Solve eqn (1) and eqn (2) and find λ = 2/3 and μ = 0
Find the coordinates of point P by putting the value of λ as (2/3, -2/3, 2/3)
Find the coordinates of point Q by putting the value of μ as (1, -1, 0)
By using distance formula find |PQ| = $\sqrt{\frac{2}{3}}\:\:units$
Question 19
An insect is crawling along the line   $\vec{r}=6\hat{i}+2\hat{j}+2\hat{k}+\lambda(\hat{i}-2\hat{j}+2\hat{k})$ and another insect is crawling along the line $\vec{r}=-4\hat{i}-\hat{k}+\mu(3\hat{i}-2\hat{j}-2\hat{k})$ . At what points on the lines should they reach so that the distance between them is the shortest? Find the shortest possible distance between them.
Answer:  λ = -1, μ =1 and shortest distance = 9
Question 20
The equation of line is
5x - 3 = 15y + 7 = 3 - 10z. Find the direction cosines of the line
Solution Hint
Find the LCM of the coefficients of x, y and z
Divide all the terms of the line by 15 and convert the given line in standard form.
Find the direction ratios as 6, 2, - 3
Question 21
Find the direction cosines from these direction ratios as (6/7, 2/7, -3/7)
Show that the following lines are skew lines
$\vec{r}=2\hat{i}-\hat{k}+\lambda (-2\hat{j}+\hat{k})$
$\vec{r}=\hat{i}+3\hat{j}+2\hat{k}+\mu(\hat{i}-2\hat{k})$
Solution Hint
If $(\overrightarrow{a}_{2}-\overrightarrow{a}_{1}).(\vec{b}_{1}\times\vec{b}_{2})\neq 0$ then lines are not intersecting.
If    $\frac{a_{1}}{a_{2}}\neq\frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}$ then the lines are not parallel.
Lines which are neither intersecting nor parallel then the lines are called skew lines.

PLANE

This topic is deleted from CBSE syllabus

Question 1: Find the distance of the given plane from the origin:
$\vec{r}.(2\hat{i}+3\hat{j}-6\hat{k})+2=0$
Answer : 2/7
Question 2: Find the angle between the planes $\vec{r}.(\hat{i}-2\hat{j}-2\hat{k})=1$
$\vec{r}.(3\hat{i}-6\hat{j}+2\hat{k})=0$
Answer   $cos^{-1}\left ( \frac{11}{21} \right )$
Question 3: Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of planes x + 2y + 3z - 4 = 0 and 2x + y - z + 5 = 0
Answer : 51x + 15y - 50z + 173 = 0 , λ = -29/7
Question 4: Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the plane x + 2y + 3z = 5 and 3x + 3y + z = 0.
Ans : 7x - 8y + 3z + 25 = 0
Question 5: Find the equation of the plane through (2, 1, -1) and (-1, 3, 4) and perpendicular to the plane x -2y + 4z = 10.
Ans : 18x + 17y + 4z -49 = 0
Question 6: Find the distance of the point (1, -2, 3) from the plane x - y + z = 5 measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$
Ans : λ = 1/7, AB = 1
Question 7: Find the distance of the point (-2, 3, -4) from the given line measured ∥ to the plane 4x + 12y - 3z + 1 = 0,   $\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3z+4}{5}$
Ans : λ = 2/3 , Point is (4, 5/2, 2), Distance = 17/2
Question 8: Find the equation of the plane passing through the point P(4, 6, 2) and the point of intersection of the plane x + y - z = 8 and the line $\frac{x-1}{3}=\frac{y}{2}= \;\frac{z+1}{7}$
Ans $\frac{x-4}{1}=\frac{y-6}{1}=\frac{z-2}{2}$
Question 9: Find the vector equation of the plane through the line of intersection of planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0. Hence find whether plane thus obtained contains the line
$\frac{x+2}{5}=\frac{y-3}{4}= \;\frac{z}{5}$
Ans 9 : $\vec{r}\left ( \hat{i}-\hat{k} \right )+2=0,$ Yes
Question 10: Find the angle between the following line and plane :
$\vec{r}=\left (\hat{i}-\hat{j}+\hat{k} \right )+\lambda (2\hat{i}-\hat{j}+3\hat{k})$ and plane   $\vec{r}.\left (2\hat{i}+\hat{j}-\hat{k} \right )=4$

Ans : 0^o

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