Solveeit Logo
Login

Question

Mathematics Question on Vectors

The equation of the line in vector and cartesian form that passes through the point with position vector 2i^−j^+4k^2\hat{i}-\hat{j}+4\hat{k} and is in the direction i^−2j^+k^\hat{i}-2\hat{j}+\hat{k} are

A

r⃗=(2i^−j^+4k^)+λ(i^+2j^−k^)\vec{r}=\left(2\hat{i}-\hat{j}+4\hat{k}\right)+\lambda\left({\hat{i}+2\hat{j}-\hat{k}}\right); x−12=y−2−1=z+1−4\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z+1}{-4}

B

r⃗=(i^+2j^−k^)+λ(2i^−j^+4k^)\vec{r}=\left(\hat{i}+2\hat{j}-\hat{k}\right)+\lambda\left(2\hat{i}-\hat{j}+4\hat{k}\right); x−12=y−2−1=z+1−4\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z+1}{-4}

C

r⃗=(i^+2j^−k^)+λ(2i^−j^+4k^)\vec{r}=\left(\hat{i}+2\hat{j}-\hat{k}\right)+\lambda\left(2\hat{i}-\hat{j}+4\hat{k}\right); x−21=y−12=z−4−1\frac{x-2}{1}=\frac{y-1}{2}=\frac{z-4}{-1}

D

r⃗=(2i^−j^+4k^)+λ(i^+2j^−k^);x−21=y−12=z−4−1\vec{r}=\left(2\hat{i}-\hat{j}+4\hat{k}\right)+\lambda \left(\hat{i}+2\hat{j}-\hat{k}\right); \frac{x-2}{1}=\frac{y-1}{2}=\frac{z-4}{-1}

Answer

r⃗=(2i^−j^+4k^)+λ(i^+2j^−k^);x−21=y−12=z−4−1\vec{r}=\left(2\hat{i}-\hat{j}+4\hat{k}\right)+\lambda \left(\hat{i}+2\hat{j}-\hat{k}\right); \frac{x-2}{1}=\frac{y-1}{2}=\frac{z-4}{-1}

Explanation

Solution

It is known that a line passing through a point with position vector a⃗\vec{a} and parallel to b⃗\vec{b} i.e., in direction of b⃗\vec{b} is given by the equation, r⃗=a⃗+λb⃗\vec{r}=\vec{a}+\lambda\vec{b}. ∴r⃗=2i^−j^+4k^+λ(i^+2j^−k^)\therefore \vec{r}=2\hat{i}-\hat{j}+4\hat{k}+\lambda\left(\hat{i}+2\hat{j}-\hat{k}\right) This is the required equation of the line in vector form. Let r⃗=xi^+yj^+zk^\vec{r}=x\hat{i}+y\hat{j}+z\hat{k} ∴xi^+yj^+zk^=(λ+2)+i^+(2λ−1)j^+(−λ+4)k^\therefore x\hat{i}+y\hat{j}+z\hat{k}=\left(\lambda+2\right)+\hat{i}+\left(2\lambda-1\right)\hat{j}+\left(-\lambda+4\right)\hat{k} Eliminating λ\lambda, we obtain the cartesian form of equation as x−21=y+12=z−4−1\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}