17. Find the values of p so that the lines <math><mrow><mfrac><mrow><mn>1</mn><mo>−</mo><mi>x</mi></mrow><mrow><mn>3</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>7</mn><mi>y</mi><mo>−</mo><mn>1</mn><mn>4</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mi>p</mi><mo>=</mo><mfrac><mrow><mi>z</mi><mo>−</mo><mn>3</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mi>a</mi><mi>n</mi><mi>d</mi><mfrac><mrow><mn>7</mn><mo>−</mo><mn>7</mn><mi>x</mi></mrow><mrow><mn>3</mn><mi>p</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>y</mi><mo>−</mo><mn>5</mn></mrow><mrow><mn>1</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>6</mn><mo>−</mo><mi>z</mi></mrow><mrow><mn>5</mn></mrow></mfrac></mrow></math> are at right angles.

1−x3=7y−142ρ=z−32and7−7x3ρ=y−51=6−z5The 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 geta1=−3,b1=2ρ7,c1=2a2=−3ρ7,b2=1,c2=−5Now, both the lines are at right anglesSo, 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

Ask & Answer Question

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

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.

3 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

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

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

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

0 Upvotes 0 Downvotes

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)

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

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

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

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

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

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

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

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

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

$c o s θ = \frac{- 3}{2 \sqrt[]{3}}$

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

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

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

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

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

$∴ | λ | = 1$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

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.

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

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 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

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.