<p><strong>49. Find the equation of the plane passing through the point <math><mrow><mrow><mo>(</mo><mrow><mo>−</mo><mn>1</mn><mo>,</mo><mn>3</mn><mo>,</mo><mn>2</mn></mrow><mo>)</mo></mrow></mrow></math> and perpendicular to each of the planes <math><mrow><mi>x</mi><mo> </mo><mo>+</mo><mn>2</mn><mi>y</mi><mo> </mo><mo>+</mo><mn>3</mn><mi>z</mi><mo> </mo><mo>=</mo><mn>5</mn><mi>a</mi><mi>n</mi><mi>d</mi><mn>3</mn><mi>x</mi><mo> </mo><mo>+</mo><mn>3</mn><mi>y</mi><mo> </mo><mo>+</mo><mo> </mo><mi>z</mi><mo> </mo><mo>=</mo><mn>0</mn><mo>.</mo></mrow></math></strong></p>

The equation of the plane passing through the point a (x +1)+ b (y −3)+ c (z −2)=0…(1) where, a, b, c are the direction ratios of normal to the plane.It is known that two planes, a1x+b1y+c1z+d1=0&a2x+b2y+c2z+d2=0 are perpendicular, if a1a2+b1b2+c1c2=0Plane (1) is perpendicular to the plane, x +2y +3z =5a.1 + b .2 + c.3 = 0a + 2b + 3c = 0 ....(2)Also, plane (1) is perpendicular to the plane, 3x +3y + z =0a .3 + b.3 + c.1 = 03a + 3b + c = 0 .....(3)From equations (2) and (3), we obtaina2×1−3×3=b3×3−1×1=c1×3−2×3⇒a−7=b8=c−3=k(say)⇒a=−7k,b=8k,c=−3kSubstituting the values of a, b, and c in equation (1), we obtain−7k(x+1)+8k(y−3)−3k(z−2)=0⇒(−7x−7)+(8y−24)−3z+6=0⇒−7x+8y−3z−25=0⇒7x−8y+3z+25=0This is the required equation of the plane.

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

49. Find the equation of the plane passing through the point $(- 1, 3, 2)$ and perpendicular to each of the planes $x + 2 y + 3 z = 5 a n d 3 x + 3 y + z = 0 .$

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Vishal Baghel | Contributor-Level 10

4 months ago

The equation of the plane passing through the point $a (x + 1) + b (y - 3) + c (z - 2) = 0 \dots (1)$ where, a, b, c are the direction ratios of normal to the plane.

It is known that two planes, $a_{1} x + b_{1} y + c_{1} z + d_{1} = 0 & a_{2} x + b_{2} y + c_{2} z + d_{2} = 0$ are perpendicular, if $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

Plane (1) is perpendicular to the plane, $x + 2 y + 3 z = 5$

$\begin{array}{l} a . 1 + b . 2 + c . 3 = 0 & a + 2 b + 3 c = 0 . . . . (2) \end{array}$

Also, plane (1) is perpendicular to the plane, $3 x + 3 y + z = 0$

$\begin{array}{l} a . 3 + b . 3 + c . 1 = 0 & 3 a + 3 b + c = 0 . . . . . (3) \end{array}$

From equations (2) and (3), we obtain

$\begin{array}{l} \frac{a}{2 \times 1 - 3 \times 3} = \frac{b}{3 \times 3 - 1 \times 1} = \frac{c}{1 \times 3 - 2 \times 3} \\ \Rightarrow \frac{a}{- 7} = \frac{b}{8} = \frac{c}{- 3} = k (s a y) \\ \Rightarrow a = - 7 k, b = 8 k, c = - 3 k \end{array}$

Substituting the values of a, b, and c in equation (1), we obtain

$\begin{array}{l} - 7 k (x + 1) + 8 k (y - 3) - 3 k (z - 2) = 0 \\ \Rightarrow (- 7 x - 7) + (8 y - 24) - 3 z + 6 = 0 \\ \Rightarrow - 7 x + 8 y - 3 z - 25 = 0 \\ \Rightarrow 7 x - 8 y + 3 z + 25 = 0 \end{array}$

This is the required equation of the 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)

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

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

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

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

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

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

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

$c o s θ = \frac{- 48}{\sqrt[]{192} \cdot 4}$

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

$c o s θ = \frac{- 3}{2 \sqrt[]{3}}$

$c o s θ = \frac{- \sqrt[]{3}}{2}$ $θ = c o s^{- 1} (\frac{- \sqrt[]{3}}{2})$

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)

$= 8 λ + 5 = \frac{8}{10} + 5 = 5.8$

If (α, β, γ) is mirror image of the point (2, 3, 4) with respect to the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ Then 2α + 3β + 4γ is

alok kumar singh

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

x = 2λ + 1, y = 3λ + 2, z = 4λ + 3

$\vec{A B}$ = (α − 2) $\hat{i}$ + (β − 3) $\hat{j}$ + (γ − 4) $\hat{k}$

Now,

(α − 2) ⋅ 2 + (β − 3) ⋅3 + (γ − 4) ⋅ 4 = 0

2α − 4 + 3β − 9 + 4γ −16 = 0

⇒ 2α + 3β + 4γ = 29

Let $λ$ be an integer. IF the shortest distance between the lines $x - λ = 2 y - 1 = - 2 z$ and $x = y + 2 λ = z - λ i s \frac{\sqrt[]{7}}{2 \sqrt[]{2}}$ , then the value of $| λ | i s . . . . . . . . . . . . . . . . . . .$

Vishal Baghel

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

$∴ S D = \frac{| - 2 λ + 3 (2 λ + \frac{1}{2}) + λ |}{\sqrt[]{14}} = \frac{| 5 λ + \frac{3}{2} |}{\sqrt[]{14}}$

$5 λ + \frac{3}{2} = - \frac{7}{2} \Rightarrow 5 λ = - 5 \Rightarrow λ = - 1$

$∴ | λ | = 1$

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49. Find the equation of the plane passing through the point $(- 1, 3, 2)$ and perpendicular to each of the planes $x + 2 y + 3 z = 5 a n d 3 x + 3 y + z = 0 .$

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49. Find the equation of the plane passing through the point (−1,3,2) and perpendicular to each of the planes x +2y +3z =5and3x +3y + z =0.

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49. Find the equation of the plane passing through the point $(- 1, 3, 2)$ and perpendicular to each of the planes $x + 2 y + 3 z = 5 a n d 3 x + 3 y + z = 0 .$