Find the equation of the plane which contains the line of intersection of the planes r → · i ^ + 2 j ^ + 3 k ^ - 4 = 0 , r → · 2 i ^ + j ^ - k ^ + 5 = 0 and which is perpendicular to the plane r → · 5 i ^ + 3 j ^ - 6 k ^ + 8 = 0 .

The given equations of the planes are r → · i ^ + 2 j ^ + 3 k ^ - 4 = 0 …(i) r → · 2 i ^ + j ^ - k ^ + 5 = 0 …(ii) The equation of the plane passing through the line intersection of the plane given in equation (i) and equation (ii) is r → · i ^ + 2 j ^ + 3 k ^ - 4 + λ r → · 2 i ^ + j ^ - k ^ + 5 = 0 , where λ is any nonzero constant. r → · 2 λ + 1 i ^ + λ + 2 j ^ + 3 - λ k ^ + ( 5 λ - 4 ) = 0 …(iii) Let, n → 1 = 2 λ + 1 i ^ + λ + 2 j ^ + 3 - λ k ^ The plane in equation (iii) is perpendicular to the plane, r → · 5 i ^ + 3 j ^ - 6 k ^ + 8 = 0 Let, n → 2 = 5 i ^ + 3 j ^ - 6 k ^ We know that two planes are perpendicular if the dot product of their normals is zero. ∴ 5 ( 2 λ + 1 ) + 3 ( λ + 2 ) - 6 ( 3 - λ ) = 0 ⇒ 19 λ - 7 = 0 ⇒ λ = 7 19 Substituting λ = 7 19 in equation (iii), we get ⇒ r → · 33 19 i ^ + 45 19 j ^ + 50 19 k ^ - 41 19 = 0 ⇒ r → · 33 i ^ + 45 j ^ + 50 k ^ - 41 = 0 …(iv) This is the vector equation of the required plane. The Cartesian equation of this plane can be obtained by substituting r → = x i ^ + y j ^ + z k ^ in equation (iv). x i ^ + y j ^ + z k ^ · 33 i ^ + 45 j ^ + 50 k ^ - 41 = 0 ⇒ 33 x + 45 y + 50 z - 41 = 0 Therefore, 33 x + 45 y + 50 z - 41 = 0 is the required equation of the plane.

Find the distance of the point - 1 , - 5 , - 10 from the point of intersection of the line r → = 2 i ^ - j ^ + 2 k ^ + λ 3 i ^ + 4 j ^ + 2 k ^ and the plane r → · i ^ - j ^ + k ^ = 5 .

The equation of the given line is r → = 2 i ^ - j ^ + 2 k ^ + λ 3 i ^ + 4 j ^ + 2 k ^ …(i) The equation of the given plane is r → · i ^ - j ^ + k ^ = 5 …(ii) Substituting the value of r → from equation (i) in equation (ii), we get 2 i ^ - j ^ + 2 k ^ + λ 3 i ^ + 4 j ^ + 2 k ^ · i ^ - j ^ + k ^ = 5 ⇒ 3 λ + 2 i ^ + 4 λ - 1 j ^ + 2 λ + 2 k ^ · i ^ - j ^ + k ^ = 5 ⇒ 3 λ + 2 - 4 λ - 1 + 2 λ + 2 = 5 ⇒ λ = 0 Substituting this value in equation (i), we get the equation of the line as r → = 2 i ^ - j ^ + 2 k ^ This means that the position vector of the point of intersection of the line and the plane is r → = 2 i ^ - j ^ + 2 k ^ This shows that the point of intersection of the given line and plane is given by the coordinate, 2 , - 1 , 2 . The distance d between the points 2 , - 1 , 2 and - 1 , - 5 , - 10 , is d = - 1 - 2 2 + - 5 + 1 2 + - 10 - 2 2 = 9 + 16 + 144 = 169 = 13 units Therefore, the distance of the given point is 13 units.

E M B I B E

MATHEMATICS PART II : Class XII>Chapter 5 - Three Dimensional Geometry>Miscellaneous Exercise on Chapter 11

Miscellaneous Exercise on Chapter 11

Author:Jharkhand Board

Jharkhand Board Mathematics Solutions for Exercise - Miscellaneous Exercise on Chapter 11

Simple step-by-step solutions to Miscellaneous Exercise on Chapter 11 questions of Three Dimensional Geometry from MATHEMATICS PART II : Class XII. Also get 3D topic explainers, cheat sheets, and unlimited doubts solving on EMBIBE.

Questions from Miscellaneous Exercise on Chapter 11 with Hints & Solutions

MEDIUM

12th Jharkhand Board

IMPORTANT

MATHEMATICS PART II : Class XII>Chapter 5 - Three Dimensional Geometry>Miscellaneous Exercise on Chapter 11>Q 16

If $O$ be the origin and the coordinates of $P$ be $(1, 2, - 3),$ then find the equation of the plane passing through $P$ and perpendicular to $O P .$

MEDIUM

12th Jharkhand Board

IMPORTANT

MATHEMATICS PART II : Class XII>Chapter 5 - Three Dimensional Geometry>Miscellaneous Exercise on Chapter 11>Q 17

Find the equation of the plane which contains the line of intersection of the planes $\vec{r} \cdot (\hat{i} + 2 \hat{j} + 3 \hat{k}) - 4 = 0, \vec{r} \cdot (2 \hat{i} + \hat{j} - \hat{k}) + 5 = 0$ and which is perpendicular to the plane $\vec{r} \cdot (5 \hat{i} + 3 \hat{j} - 6 \hat{k}) + 8 = 0$ .

MEDIUM

12th Jharkhand Board

IMPORTANT

MATHEMATICS PART II : Class XII>Chapter 5 - Three Dimensional Geometry>Miscellaneous Exercise on Chapter 11>Q 18

Find the distance of the point $(- 1, - 5, - 10)$ from the point of intersection of the line $\vec{r} = 2 \hat{i} - \hat{j} + 2 \hat{k} + λ (3 \hat{i} + 4 \hat{j} + 2 \hat{k})$ and the plane $\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5 .$

MEDIUM

12th Jharkhand Board

IMPORTANT

MATHEMATICS PART II : Class XII>Chapter 5 - Three Dimensional Geometry>Miscellaneous Exercise on Chapter 11>Q 19

Find the vector equation of the line passing through $(1, 2, 3)$ and parallel to the planes $\vec{r} . (\hat{i} - \hat{j} + 2 \hat{k}) = 5$ and $\vec{r} \cdot (3 \hat{i} + \hat{j} + \hat{k}) = 6$ .

MEDIUM

12th Jharkhand Board

IMPORTANT

MATHEMATICS PART II : Class XII>Chapter 5 - Three Dimensional Geometry>Miscellaneous Exercise on Chapter 11>Q 20

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 $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}$ .

MEDIUM

12th Jharkhand Board

IMPORTANT

MATHEMATICS PART II : Class XII>Chapter 5 - Three Dimensional Geometry>Miscellaneous Exercise on Chapter 11>Q 21

Prove that if a plane has the intercepts $a, b, c$ and is at a distance of $p$ units from the origin, then $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$ .