The Cartesian equation of a line is 6x-2=3y+1=2z-2. Then its vector from is: Option 1: r=(2i-j+2k)+(6i+3j+2k) Option 2: r=(i+2j+3k)+(1/3i-1/3j+k) Option 3: r=(-1/3i+1/3j-k)+(i+2j+3k) Option 4: r=(1/3i-1/3j+k)+(i+2j+3k)

The answer is 'r=(1/3i-1/3j+k)+(i+2j+3k)'. View full solution:

CUET Mathematics - The Cartesian equation of a line is 6x-2=3y+1=2z-2. Then its vector from is: | PYQs + Solutions

CUET Mathematics 2026 30th May Shift 1

Geometry

Medium

core

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

$\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda\left(\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}+\hat{k}\right)$

$\vec{r}=\left(-\frac{1}{3}\hat{i}+\frac{1}{3}\hat{j}-\hat{k}\right)+\lambda(\hat{i}+2\hat{j}+3\hat{k})$

$\vec{r}=\left(\frac{1}{3}\hat{i}-\frac{1}{3}\hat{j}+\hat{k}\right)+\lambda(\hat{i}+2\hat{j}+3\hat{k})$

✅ Correct Option: 4