35. Find the angle between the planes whose vector equations are r→.(2i^+2j^−3k^)=5&r→.(3i^−3j^+5k^)=3

The equations of the given planes are r→.(2i^+2j^−3k^)=5&r→.(3i^−3j^+5k^)=3 It is known that if n1 and n2 are normal to the planes, r→.n1→=d1&r→.n2→=d2 then the angle between them, Q, is given by, cosQ=|n1→.n2→|n1→||n2→||..........(1) H e r e , n 1 → = 2 i ^ + 2 j ^ − 3 k ^ & n 2 → = 3 i ^ − 3 j ^ + 5 k ^ ∴ n 1 → . n 2 → = ( 2 i ^ + 2 j ^ − 3 k ^ ) ( 3 i ^ − 3 j ^ + 5 k ^ ) = 2 . 3 + 2 . ( − 3 ) + ( − 3 ) . 5 = − 1 5 | n 1 → | = = | n 2 → | = = Substituting the value of n1→.n2→ |n1→|&|n2→| in equation (1), we obtain cosQ=|−15.|⇒cosQ=15⇒cosQ−1=(15)

35. Find the angle between the planes whose vector equations are $\vec{r} . (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 5 & \vec{r} . (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 3$

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Vishal Baghel | Contributor-Level 10

10 months ago

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

It is known that if n1 and n2 are normal to the planes, $\vec{r} . \vec{n_{1}} = d_{1} & \vec{r} . \vec{n_{2}} = d_{2}$ then the angle between them, Q, is given by,

$c o s Q = | \frac{\vec{n_{1}} . \vec{n_{2}}}{| \vec{n_{1}} | | \vec{n_{2}} |} | . . . . . . . . . . (1)$

$\begin{array}{l} H e r e, \vec{n_{1}} = 2 \hat{i} + 2 \hat{j} - 3 \hat{k} & \vec{n_{2}} = 3 \hat{i} - 3 \hat{j} + 5 \hat{k} \\ ∴ \vec{n_{1}} . \vec{n_{2}} = (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 2.3 + 2 . (- 3) + (- 3) . 5 = - 15 \\ | \vec{n_{1}} | = = \\ | \vec{n_{2}} | = = \end{array}$

Substituting the value of $\vec{n_{1}} . \vec{n_{2}}$ $| \vec{n_{1}} | & | \vec{n_{2}} |$ in equation (1), we obtain

$\begin{array}{l} c o s Q = | \frac{- 15}{.} | \\ \Rightarrow c o s Q = \frac{15}{} \\ \Rightarrow c o s Q^{- 1} = (\frac{15}{}) \end{array}$

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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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