The point having position vector vec{p} = 4ha | vedclass

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

Question

(D) point $\vec{p}$ lies on the line $\vec{r} = \vec{a} + \lambda\vec{b}$ if $\vec{p} - \vec{a}$ is parallel to $\vec{b}$, i.e., $\vec{p} - \vec{a} =

Vedclass · Accepted Answer

$\vec{r} = (6\hat{i} - 4\hat{j} + 5\hat{k}) + \lambda(2\hat{i} + 7\hat{j} + 3\hat{k})$

Vedclass · Answer

(D) point $\vec{p}$ lies on the line $\vec{r} = \vec{a} + \lambda\vec{b}$ if $\vec{p} - \vec{a}$ is parallel to $\vec{b}$, i.e., $\vec{p} - \vec{a} = k\vec{b}$ for some scalar $k$.Given $\vec{p} = 4\hat{i} - 11\hat{j} + 2\hat{k}$ and $\vec{a} = 6\hat{i} - 4\hat{j} + 5\hat{k}$.$\vec{p} - \vec{a} = (4-6)\hat{i} + (-11 - (-4))\hat{j} + (2-5)\hat{k} = -2\hat{i} - 7\hat{j} - 3\hat{k}$.For option $(D)$, $\vec{b} = 2\hat{i} + 7\hat{j} + 3\hat{k}$.Since $\vec{p} - \vec{a} = -1(2\hat{i} + 7\hat{j} + 3\hat{k}) = -1\vec{b}$, the vector $\vec{p} - \vec{a}$ is parallel to $\vec{b}$.Thus, the point lies on the line given in option $(D)$.

The point having position vector $\vec{p} = 4\hat{i} - 11\hat{j} + 2\hat{k}$ lies on which of the following lines?

Similar Questions

If for some $m \in \mathbb{R}$ the lines $L_1 : \frac{x + 1}{m} = \frac{y - m}{-1} = \frac{z - 1}{1}$ and $L_2 : \frac{x + 2}{-4} = \frac{y + 1}{9} = \frac{z + 1}{1}$ are coplanar, then line $L_1$ passes through the point

Let $P$ be the image of the point $Q(7,-2,5)$ in the line $L: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $R(5, p, q)$ be a point on $L$. Then the square of the area of $\triangle P Q R$ is $\qquad$

If the foot of the perpendicular from point $(4,3,8)$ on the line $L_{1}: \frac{x-a}{l}=\frac{y-2}{3}=\frac{z-b}{4},$ $l \neq 0$ is $(3,5,7),$ then the shortest distance between the line $L_{1}$ and line $L_{2}: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is equal to:

The shortest distance between the skew lines $\vec{r}=(-\hat{i}-2 \hat{j}-3 \hat{k})+t(3 \hat{i}-2 \hat{j}-2 \hat{k})$ and $\vec{r}=(7 \hat{i}+4 \hat{k})+s(\hat{i}-2 \hat{j}+2 \hat{k})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module