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MHT CET · Maths · Three Dimensional Geometry

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. A \(\bar{r}=(-2 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}+5 \hat{k})\)
  2. B \(\bar{r}=(2 \hat{i}-4 \hat{j}-5 \hat{k})+\lambda(-3 \hat{i}+2 \hat{j}-5 \hat{k})\)
  3. C \(\bar{r}=(-2 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda(10 \hat{i}+25 \hat{j}-16 \hat{k})\)
  4. D \(\bar{r}=(3 \hat{i}+2 \hat{j}+5 \hat{k})+\lambda(10 \hat{i}+25 \hat{j}-16 \hat{k})\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\bar{r}=(2 \hat{i}-4 \hat{j}-5 \hat{k})+\lambda(-3 \hat{i}+2 \hat{j}-5 \hat{k})\)

Step-by-step Solution

Detailed explanation

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