The Cartesian equation of a line is $\frac{x+2}{3}=\frac{y-4}{2}=\frac{z-5}{5}$, then the vector equation of…

The Cartesian equation of a line is $\frac{x+2}{3}=\frac{y-4}{2}=\frac{z-5}{5}$, then the vector equation of the line is
  1. $\bar{r}=(-2 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}+5 \hat{k})$
  2. $\bar{r}=(2 \hat{i}-4 \hat{j}-5 \hat{k})+\lambda(-3 \hat{i}+2 \hat{j}-5 \hat{k})$
  3. $\bar{r}=(-2 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda(10 \hat{i}+25 \hat{j}-16 \hat{k})$
  4. $\bar{r}=(3 \hat{i}+2 \hat{j}+5 \hat{k})+\lambda(10 \hat{i}+25 \hat{j}-16 \hat{k})$

Solution

The line $\frac{x+2}{3}=\frac{y-4}{2}=\frac{z-5}{5}$ is passing through $(-2,4,5)$ and has d.r's $ < 3,2,5>$ Hence, the vector equation is $\vec{r}=(-2 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}+5 \hat{k})$

Asked in: MHT CET 2022 (08 Aug Shift 1)

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