Let L be the line parallel to the vector sqrt | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equation of the line $L$ passing through point $A(\hat{i}+2 \hat{j}-3 \hat{k})$ and parallel to vector $\vec{v} = \sqrt{2} \hat{i}-5 \hat{j}+3

Vedclass · Accepted Answer

$(1-3 \sqrt{2}) \hat{i}+17 \hat{j}-12 \hat{k}$

Vedclass · Answer

(A) The equation of the line $L$ passing through point $A(\hat{i}+2 \hat{j}-3 \hat{k})$ and parallel to vector $\vec{v} = \sqrt{2} \hat{i}-5 \hat{j}+3 \hat{k}$ is given by $\vec{r} = (\hat{i}+2 \hat{j}-3 \hat{k}) + \lambda(\sqrt{2} \hat{i}-5 \hat{j}+3 \hat{k})$.Any point $P$ on the line $L$ can be represented as $P = (\sqrt{2} \lambda+1, -5 \lambda+2, 3 \lambda-3)$.The distance $AP$ is given as $18$ units.$AP^2 = (\sqrt{2} \lambda+1-1)^2 + (-5 \lambda+2-2)^2 + (3 \lambda-3+3)^2 = 18^2$.$2 \lambda^2 + 25 \lambda^2 + 9 \lambda^2 = 324$.$36 \lambda^2 = 324$.$\lambda^2 = 9$, which gives $\lambda = \pm 3$.For $\lambda = 3$, $P = (3\sqrt{2}+1)\hat{i} - 13\hat{j} + 6\hat{k}$.For $\lambda = -3$, $P = (-3\sqrt{2}+1)\hat{i} + 15+2\hat{j} - 9-3\hat{k} = (1-3\sqrt{2})\hat{i} + 17\hat{j} - 12\hat{k}$.Comparing with the options, the correct position vector is $(1-3 \sqrt{2}) \hat{i}+17 \hat{j}-12 \hat{k}$.

Let $L$ be the line parallel to the vector $\sqrt{2} \hat{i}-5 \hat{j}+3 \hat{k}$ and passing through the point $A$ given by $\hat{i}+2 \hat{j}-3 \hat{k}$. If the distance between $A$ and a point $P$ on the line $L$ is $18$ units, then the position vector of such a point $P$ is

Similar Questions

If the square of the shortest distance between the lines $\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}$ and $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}$ is $\frac{m}{n}$,where $m, n$ are coprime numbers,then $m+n$ is equal to:

The angle between the lines $\frac{x-1}{l}=\frac{y+1}{m}=\frac{z}{n}$ and $\frac{x+1}{m}=\frac{y-3}{n}=\frac{z-1}{l}$,where $l > m > n$ and $l, m, n$ are roots of the equation $x^3+x^2-4x-4=0$,is

The equation of the line passing through the points $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ is given by:

The vector equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$ is . . . . . . .

Find the position vector of the image of the point with position vector $\vec{P} = 2\hat{i} + \hat{j} + 3\hat{k}$ in the line whose vector equation is $\vec{r} = \hat{j} - 2\hat{k} + \lambda(\hat{i} + \hat{j} - \hat{k})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module