17. Find the values of p so that the lines 1−x3=7y−142p=z−32and7−7x3p=y−51=6−z5 are at right angles.

1−x3=7y−142ρ=z−32and7−7x3ρ=y−51=6−z5 The standard form of a pair of Cartesian lines is; x−x1a1=y−y1b1=z−z1c1andx−x2a2=y−y2b2=z−z2c2−−−−−(1) So, −(x−1)3=7(y−5)2ρ=z−32and−7(x−1)3ρ=y−51=−(z−6)5x−1−3=y−22ρ/7=z−32andx−1−3ρ/7=y−51=z−6−5−−−−−(2) Comparing (1) and (2) we get a1=−3,b1=2ρ7,c1=2a2=−3ρ7,b2=1,c2=−5 Now, both the lines are at right angles So, a1a2+b1b2+c1c2=0 ⇒(−3)*(−3ρ)7+2ρ7*1+2*(−5)=0⇒9ρ7+2ρ7+(−10)=0⇒9ρ+2ρ7=10⇒11ρ=70ρ=7011 ∴ The value of ρ is 7011

17. Find the values of p so that the lines $\frac{1 - x}{3} = \frac{7 y - 14}{2} p = \frac{z - 3}{2} a n d \frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}$ are at right angles.

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Vishal Baghel | Contributor-Level 10

10 months ago

$\frac{1 - x}{3} = \frac{7 y - 14}{2 ρ} = \frac{z - 3}{2} a n d \frac{7 - 7 x}{3 ρ} = \frac{y - 5}{1} = \frac{6 - z}{5}$

The standard form of a pair of Cartesian lines is;

$\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}} a n d \frac{x - x_{2}}{a_{2}} = \frac{y - y_{2}}{b_{2}} = \frac{z - z_{2}}{c_{2}} - - - - - (1)$

So,

$\begin{array}{l} \frac{- (x - 1)}{3} = \frac{7 (y - 5)}{2 ρ} = \frac{z - 3}{2} a n d \frac{- 7 (x - 1)}{3 ρ} = \frac{y - 5}{1} = \frac{- (z - 6)}{5} \\ \frac{x - 1}{- 3} = \frac{y - 2}{2 ρ / 7} = \frac{z - 3}{2} a n d \frac{x - 1}{- 3 ρ / 7} = \frac{y - 5}{1} = \frac{z - 6}{- 5} - - - - - (2) \end{array}$

Comparing (1) and (2) we get

$\begin{array}{l} a_{1} = - 3, b_{1} = \frac{2 ρ}{7}, c_{1} = 2 \\ a_{2} = \frac{- 3 ρ}{7}, b_{2} = 1, c_{2} = - 5 \end{array}$

Now, both the lines are at right angles

So, $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

$\begin{array}{l} \Rightarrow (- 3) * \frac{(- 3 ρ)}{7} + \frac{2 ρ}{7} * 1 + 2 * (- 5) = 0 \\ \Rightarrow \frac{9 ρ}{7} + \frac{2 ρ}{7} + (- 10) = 0 \\ \Rightarrow \frac{9 ρ + 2 ρ}{7} = 10 \\ \Rightarrow 11 ρ = 70 \\ ρ = \frac{70}{11} \end{array}$

$∴$ The value of $ρ$ is $\frac{70}{11}$

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

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

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$= \frac{π}{4} (π + α) - 2$

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

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

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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)

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

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

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

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

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

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

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

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Also, P lie on line

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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

$λ = \frac{1}{10}$

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Vishal Baghel

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