10. Find the equation of the line in vector and in Cartesian form that passes through the point with position vector <math><mrow><mn>2</mn><mover accent="true"><mi>i</mi><mo>^</mo></mover><mo>+</mo><mover accent="true"><mi>j</mi><mo>^</mo></mover><mo>+</mo><mn>4</mn><mover accent="true"><mi>k</mi><mo>^</mo></mover></mrow></math> and is in the direction <math><mrow><mover accent="true"><mi>i</mi><mo>^</mo></mover><mo>+</mo><mn>2</mn><mover accent="true"><mi>j</mi><mo>^</mo></mover><mo>−</mo><mover accent="true"><mi>k</mi><mo>^</mo></mover></mrow></math>

The line passes through the point with position vector, a→=2i^−j^+4k^−−−−−(1)The given vector: b→=i^+2j^−k^−−−−−(2)The line which passes through a point with position vector a→ and parallel to b→ is given by,r→=a→+λb→r→=2i^−j^+4k^+λ(i^+2j^−k^)∴ This is required equation of the line in vector form.Now,Let r→=xi^−yj^+zk^⇒xi^−yj^+zk^=(λ+2)i^+(2λ−1)j^+(−λ+4)k^Comparing the coefficient to eliminate λ ,x=λ+2,x1=2,a=1y=2λ−1,y1=−1,b=2z=−λ+4,z1=4,c=−1x−21=y+12=z−4−1

Ask & Answer Question

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

10. Find the equation of the line in vector and in Cartesian form that passes through the point with position vector $2 \hat{i} + \hat{j} + 4 \hat{k}$ and is in the direction $\hat{i} + 2 \hat{j} - \hat{k}$

10 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

The line passes through the point with position vector, $\vec{a} = 2 \hat{i} - \hat{j} + 4 \hat{k} - - - - - (1)$

The given vector: $\vec{b} = \hat{i} + 2 \hat{j} - \hat{k} - - - - - (2)$

The line which passes through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is given by,

$\begin{array}{l} \vec{r} = \vec{a} + λ \vec{b} \\ \vec{r} = 2 \hat{i} - \hat{j} + 4 \hat{k} + λ (\hat{i} + 2 \hat{j} - \hat{k}) \end{array}$

$∴$ This is required equation of the line in vector form.

Now,

Let $\begin{array}{l} \vec{r} = x \hat{i} - y \hat{j} + z \hat{k} \\ \Rightarrow x \hat{i} - y \hat{j} + z \hat{k} = (λ + 2) \hat{i} + (2 λ - 1) \hat{j} + (- λ + 4) \hat{k} \end{array}$

Comparing the coefficient to eliminate $λ$ ,

$\begin{array}{l} x = λ + 2, x_{1} = 2, a = 1 \\ y = 2 λ - 1, y_{1} = - 1, b = 2 \\ z = - λ + 4, z_{1} = 4, c = - 1 \end{array}$

$\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{- 1}$

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

10. Find the equation of the line in vector and in Cartesian form that passes through the point with position vector $2 \hat{i} + \hat{j} + 4 \hat{k}$ and is in the direction $\hat{i} + 2 \hat{j} - \hat{k}$

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

10. Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2i^+j^+4k^ and is in the direction i^+2j^−k^

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

10. Find the equation of the line in vector and in Cartesian form that passes through the point with position vector $2 \hat{i} + \hat{j} + 4 \hat{k}$ and is in the direction $\hat{i} + 2 \hat{j} - \hat{k}$