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Ncert Solutions Maths class 12th Maths Class 12th Three Dimensional Geometry

33. Find the vector equation of the plane passing through the intersection of the planes $\vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) = 7, \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9$ and through the point $(2, 1, 3) .$

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Vishal Baghel | Contributor-Level 10

4 months ago

The equations of the planes are $\vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) = 7, \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9$

$\begin{array}{l} \Rightarrow \vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) - 7 = 0 . . . . . . . . . . (1) \\ \Rightarrow \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) - 9 = 0 . . . . . . . . . . (2) \end{array}$

The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,

$\begin{array}{l} [\vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) - 7] + λ [\vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) - 9] = 0, w h e r e, λ \in R \\ \vec{r} . [(2 \hat{i} + 2 \hat{j} - 2 \hat{k}) + λ (2 \hat{i} + 5 \hat{j} + 3 \hat{k})] = 9 λ + 7 \\ \vec{r} . [(2 + 2 λ) \hat{i} + (2 + 5 λ) \hat{j} + (3 λ - 3) \hat{k}] = 9 λ + 7 . . . . . . . . . . (3) \end{array}$

The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,

$\vec{r} = 2 \hat{i} + 1 \hat{j} + 3 \hat{k}$

Substituting in equation (3), we obtain

$\begin{array}{l} (2 \hat{i} + \hat{j} + 3 \hat{k}) . [(2 + 2 λ) \hat{i} + (2 + 5 λ) \hat{j} + (3 λ - 3) \hat{k}] = 9 λ + 7 \\ \Rightarrow 2 (2 + 2 λ) + 1 (2 + 5 λ) + 3 (3 λ - 3) = 9 λ + 7 \\ \Rightarrow 4 + 4 λ + 2 + 5 λ + 9 λ - 9 = 9 λ + 7 \\ \Rightarrow 18 λ - 3 = 9 λ + 7 \\ \Rightarrow 9 λ = 10 \\ \Rightarrow λ = \frac{10}{9} \end{array}$

Substituting $λ = \frac{10}{9}$ in equation (3), we obtain

$\begin{array}{l} \vec{r} . (\frac{38}{9} \hat{i} + \frac{68}{9} \hat{j} + \frac{3}{9} \hat{k}) = 17 \\ \Rightarrow \vec{r} . (38 \hat{i} + 68 \hat{j} + 3 \hat{k}) = 153 \end{array}$

This is the vector equation of the required plane.

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If $\int_{}^{} (\frac{x^{2} c o s x}{1 + π^{x}} + \frac{1 + s i n^{2} x}{1 + e^{{(s i n x)}^{2023}}}) d x = \frac{π}{4} (π + α) - 2$

Then the value of 'α' is equal to

alok kumar singh

$\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{x^{2} c o s x}{1 + π^{2}} + \frac{1 + s i n^{2} x}{1 + e^{{(s i n x)}^{2023}}}) d x = \frac{π}{4} (π + α) - 2$

$\int_{0}^{\frac{π}{2}} {(\frac{x^{2} c o s x}{1 + π^{x}} + \frac{1 + s i n^{2} x}{1 + e^{{(s i n x)}^{2023}}}) + (\frac{x^{2} c o s x}{1 + π^{- x}} + \frac{1 + s i n^{2} x}{1 + e^{- {(s i n x)}^{2023}}})} d x$

$= \frac{π}{4} (π + α) - 2$

$\int_{0}^{\frac{π}{2}} (x^{2} c o s x + 1 + s i n^{2} x) d x = \frac{π}{4} (π + α) - 2$

$\int_{0}^{\frac{π}{2}} x^{2} c o s x d x + \int_{0}^{\frac{π}{2}} (1 + s i n^{2} x) d x = \frac{π}{4} (π + α) - 2$ ....(1)

Let $I_{1} = \int_{0}^{\frac{π}{2}} (1 + s i n^{2} x) d x$

$I_{1} = \int_{0}^{\frac{π}{2}} 1 \cdot d x + \int_{0}^{\frac{π}{2}} (\frac{1 - c o s 2 x}{2}) d x$

$I_{1} = \frac{π}{2} + \frac{1}{2} [\frac{π}{2} + 0]$

$I_{1} = \frac{3 π}{4}$

Let $I_{2} = \int_{0}^{\frac{π}{2}} x^{2} c o s x d x$

$I_{2} = {[x^{2} (s i n x) - \int 2 x \int c o s x d x]}_{0}^{\frac{π}{2}}$

$I_{2} = {[x^{2} (s i n x) - 2 \int x s i n x]}_{0}^{\frac{π}{2}}$

$I_{2} = {[x^{2} s i n x - 2 (x (- c o s x) + \int c o s x)]}_{0}^{\frac{π}{2}}$

$I_{2} = {[x^{2} s i n x - 2 (- x c o s x + s i n x)]}_{0}^{\frac{π}{2}}$

$I_{2} = (\frac{π^{2}}{4} - 2)$

Put l₁ and l₂ in (1)

$\frac{π^{2}}{4} - 2 + \frac{3 π}{4}$

$\frac{π^{2}}{4} + \frac{3 π}{4} - 2$

$\frac{π}{4} (π + 3) - 2$

α = 3

If $| \vec{a} | = 1$ , $| \vec{b} | = 4$ $\vec{a} \cdot \vec{b} = 2$ and $\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b}$ Then the angle between $\vec{b}$ and $\vec{c}$ is

alok kumar singh

Given $| \vec{a} | = 1$ , $| \vec{b} | = 4$ , $\vec{a} \cdot \vec{b} = 2$

$\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b}$

Dot product with $\vec{a}$ on both sides

$\vec{c} \cdot \vec{a} = - 6$ ... (1)

Dot product with $\vec{b}$ on both sides

$\vec{b} \cdot \vec{c} = - 48$ ... (2)

$\vec{c} \cdot \vec{c} = 4 {| \vec{a} \times \vec{b} |}^{2} + 9 {| \vec{b} |}^{2}$

${| \vec{c} |}^{2} = 4 [{| \vec{a} |}^{2} \cdot {| \vec{b} |}^{2} - {(\vec{a} \cdot \vec{b})}^{2}] + 9 {| \vec{b} |}^{2}$

${| \vec{c} |}^{2} = 4 [(1) {(4)}^{2} - (4)] + 9 (16)$

${| \vec{c} |}^{2} = 4 [12] + 144$

${| \vec{c} |}^{2} = 48 + 144$

${| \vec{c} |}^{2} = 192$

$c o s θ = \frac{\vec{b} \cdot \vec{c}}{| \vec{b} | | \vec{c} |}$

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$c o s θ = \frac{- 48}{8 \sqrt[]{3} \cdot 4}$

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If the foot of perpendicular from (1, 2, 3) to the line $\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z - 4}{1}$ is (a, b, g) then find a + b + g

alok kumar singh

(a – 1) × 2 + (b – 2) × 5 + (g – 3) × 1 = 0

2a + 5b + g – 15 = 0

Also, P lie on line

a + 1 = 2λ

b – 2 = 5λ

g – 4 = λ

2 (2λ – 1) + 5 (5λ + 2) + λ + 4 – 15 = 0

4λ + 25λ + λ – 2 + 10 + 4 – 15 = 0

30λ – 3 = 0

$λ = \frac{1}{10}$

a + b + g = (2λ – 1) + (5λ + 2) + (λ + 4)

$= 8 λ + 5 = \frac{8}{10} + 5 = 5.8$

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Take $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = λ$

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Now,

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Vishal Baghel

$L_{1} = \frac{x - λ}{1} = \frac{y - \frac{1}{2}}{\frac{1}{2}} = \frac{z}{- \frac{1}{2}}$

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$∴ | λ | = 1$

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33. Find the vector equation of the plane passing through the intersection of the planes $\vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) = 7, \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9$ and through the point $(2, 1, 3) .$

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33. Find the vector equation of the plane passing through the intersection of the planes r→.(2i^+2j^−2k^)=7,r→.(2i^+5j^+3k^)=9 and through the point (2,1,3).

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33. Find the vector equation of the plane passing through the intersection of the planes $\vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) = 7, \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9$ and through the point $(2, 1, 3) .$