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

The equations of the given planes are r→.(i^+2j^+3k^)−4=0.....(1)r→.(2i^+j^−k^)+5=0......(2) The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is [r→.(i^+2j^+3k^)−4]+λ[r→.(2i^+j^−k^)+5]=0r→.[(2λ+1)i^+(λ+2)j^+(3−λ)k^]+(5λ−4)=0.....(3) The plane in equation (3) is perpendicular to the plane, r→.(5i^+3j^−6k^)+8=0 ∴5(2λ+1)+3(λ+2)−6(3−λ)=0⇒19λ−7=0⇒λ=719 Substituting λ = 7/19 in equation (3), we obtain ⇒r→.(1319i^+4519j^+5019k^)−4119=0⇒r→.(33i^+45j^+50k^)−41=0 This is the vector equation of the required plane. The Cartesian equation of this plane can be obtained by substituting r→=xi^+yj^+zk^ in equation (3). (xi^+yj^+zk^).(33i^+45j^+50k^)−41=0⇒33x+45y+50z−41=0

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

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Vishal Baghel | Contributor-Level 10

10 months ago

The equations of the given planes are

$\begin{array}{l} \vec{r} . (\hat{i} + 2 \hat{j} + 3 \hat{k}) - 4 = 0 . . . . . (1) \\ \vec{r} . (2 \hat{i} + \hat{j} - \hat{k}) + 5 = 0 . . . . . . (2) \end{array}$

The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is

$\begin{array}{l} [\vec{r} . (\hat{i} + 2 \hat{j} + 3 \hat{k}) - 4] + λ [\vec{r} . (2 \hat{i} + \hat{j} - \hat{k}) + 5] = 0 \\ \vec{r} . [(2 λ + 1) \hat{i} + (λ + 2) \hat{j} + (3 - λ) \hat{k}] + (5 λ - 4) = 0 . . . . . (3) \end{array}$

The plane in equation (3) is perpendicular to the plane, $\vec{r} . (5 \hat{i} + 3 \hat{j} - 6 \hat{k}) + 8 = 0$

$\begin{array}{l} ∴ 5 (2 λ + 1) + 3 (λ + 2) - 6 (3 - λ) = 0 \\ \Rightarrow 19 λ - 7 = 0 \\ \Rightarrow λ = \frac{7}{19} \end{array}$

Substituting λ = 7/19 in equation (3), we obtain

$\begin{array}{l} \Rightarrow \vec{r} . (\frac{13}{19} \hat{i} + \frac{45}{19} \hat{j} + \frac{50}{19} \hat{k}) \frac{- 41}{19} = 0 \\ \Rightarrow \vec{r} . (33 \hat{i} + 45 \hat{j} + 50 \hat{k}) - 41 = 0 \end{array}$

This is the vector equation of the requ

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Similar Questions for you

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)

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

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

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

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

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

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

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

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

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

$c o s θ = \frac{- 3}{2 \sqrt[]{3}}$

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

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)

$= 8 λ + 5 = \frac{8}{10} + 5 = 5.8$

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

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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