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 Textbook for Class XII>Chapter 5 - Three Dimensional Geometry>Miscellaneous Exercise on Chapter 11

Mizoram Board Solutions for Chapter: Three Dimensional Geometry, Exercise 4: Miscellaneous Exercise on Chapter 11

Author:Mizoram Board

Mizoram Board Mathematics Solutions for Exercise - Mizoram Board Solutions for Chapter: Three Dimensional Geometry, Exercise 4: Miscellaneous Exercise on Chapter 11

Attempt the practice questions on Chapter 5: Three Dimensional Geometry, Exercise 4: Miscellaneous Exercise on Chapter 11 with hints and solutions to strengthen your understanding. MATHEMATICS PART II Textbook for Class XII solutions are prepared by Experienced Embibe Experts.

Questions from Mizoram Board Solutions for Chapter: Three Dimensional Geometry, Exercise 4: Miscellaneous Exercise on Chapter 11 with Hints & Solutions

MEDIUM

12th Mizoram Board

IMPORTANT

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

If $1, 2$ 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 Mizoram Board

IMPORTANT

MATHEMATICS PART II Textbook for 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 Mizoram Board

IMPORTANT

MATHEMATICS PART II Textbook for 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 Mizoram Board

IMPORTANT

MATHEMATICS PART II Textbook for 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 Mizoram Board

IMPORTANT

MATHEMATICS PART II Textbook for 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 Mizoram Board

IMPORTANT

MATHEMATICS PART II Textbook for 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 $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$ .