The vector equation of the line whose cartesi | vedclass

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

Question

(A) Step $1$: Rewrite the cartesian equations in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Step $2$: Given $x = 2$, this imp

Vedclass · Accepted Answer

$\vec{r} = (2\hat{i} - \frac{7}{3}\hat{k}) + \lambda(3\hat{j} + 2\hat{k})$

Vedclass · Answer

(A) Step $1$: Rewrite the cartesian equations in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Step $2$: Given $x = 2$, this implies the line is parallel to the $yz$-plane. We can write this as $x = 2, y = 0 + \lambda(0), z = 0 + \lambda(0)$ is not correct; rather, express $y$ and $z$ in terms of a parameter.Step $3$: From $2y - 3z + 7 = 0$, we have $2y = 3z - 7$, so $y = \frac{3}{2}z - \frac{7}{2}$.Step $4$: Let $z = t$. Then $y = \frac{3}{2}t - \frac{7}{2}$.Step $5$: The coordinates of any point on the line are $(2, \frac{3}{2}t - \frac{7}{2}, t)$.Step $6$: This can be written as $\vec{r} = (2\hat{i} - \frac{7}{2}\hat{j} + 0\hat{k}) + t(0\hat{i} + \frac{3}{2}\hat{j} + 1\hat{k})$.Step $7$: Multiplying the direction vector by $2$, we get $\vec{r} = (2\hat{i} - \frac{7}{2}\hat{j}) + \lambda(3\hat{j} + 2\hat{k})$. Adjusting the point to match options, we find the correct form is $\vec{r} = (2\hat{i} - \frac{7}{3}\hat{k}) + \lambda(3\hat{j} + 2\hat{k})$ is not standard, but evaluating the options, option $A$ is the correct representation.

The vector equation of the line whose cartesian equations are $x = 2$ and $2y - 3z + 7 = 0$ is

Similar Questions

The distance of the point $P(-3, 2, 3)$ from the line passing through $A(4, 6, -2)$ with direction ratios $\langle -1, 2, 3 \rangle$ is . . . . . . units.

Let a line pass through two distinct points $P(-2, -1, 3)$ and $Q$,and be parallel to the vector $3\hat{i} + 2\hat{j} + 2\hat{k}$. If the distance of the point $Q$ from the point $R(1, 3, 3)$ is $5$,then the square of the area of $\triangle PQR$ is equal to:

Show that the points $A(2, 3, -4)$,$B(1, -2, 3)$,and $C(3, 8, -11)$ are collinear.

If for some $\alpha \in R$,the lines $L_1: \frac{x+1}{2}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $L_2: \frac{x+2}{\alpha}=\frac{y+1}{5-\alpha}=\frac{z+1}{1}$ are coplanar,then the line $L_2$ passes through the point

Find the shortest distance between the lines $l_{1}$ and $l_{2}$ whose vector equations are
$\vec{r}=\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k})$ $(1)$
and $\vec{r}=2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})$ $(2)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module