MHT CET · Maths · Three Dimensional Geometry
The Cartesian equation of a line is \(2 x-2=3 y+1=6 z-2\), then the vector equation of the line is
- A \(\bar{r}=\left(\hat{i}-\frac{\hat{j}}{3}+\frac{\hat{k}}{3}\right)+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})\)
- B \(\overline{\mathrm{r}}=\left(-\hat{\mathrm{i}}+\frac{\hat{\mathrm{j}}}{3}-\frac{\hat{\mathrm{k}}}{3}\right)+\lambda\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{6} \hat{\mathrm{k}}\right)\)
- C \(\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})\)
- D \(\overline{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{6} \hat{\mathrm{k}}\right)\)
Answer & Solution
Correct Answer
(A) \(\bar{r}=\left(\hat{i}-\frac{\hat{j}}{3}+\frac{\hat{k}}{3}\right)+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})\)
Step-by-step Solution
Detailed explanation
Given Cartesian equation of the line is
\(\begin{aligned}
& 2 x-2=3 y+1=6 z-2 \\
& \Rightarrow 2(x-1)=3\left(y+\frac{1}{3}\right)=6\left(z-\frac{1}{3}\right) \\
& \Rightarrow \frac{x-1}{\frac{1}{2}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-\frac{1}{3}}{\frac{1}{6}} \\
& \Rightarrow \frac{x-1}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{1}{3}}{1}
\end{aligned}\)
\(\therefore \quad\) The given line passes through \(\left(1,-\frac{1}{3}, \frac{1}{3}\right)\) and has direction ratios proportional to \(3,2,1\).
\(\therefore \quad\) Vector equation is \(\overline{\mathrm{r}}=\left(\hat{\mathrm{i}}-\frac{\hat{\mathrm{j}}}{3}+\frac{\hat{\mathrm{k}}}{3}\right)+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})\)
\(\begin{aligned}
& 2 x-2=3 y+1=6 z-2 \\
& \Rightarrow 2(x-1)=3\left(y+\frac{1}{3}\right)=6\left(z-\frac{1}{3}\right) \\
& \Rightarrow \frac{x-1}{\frac{1}{2}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-\frac{1}{3}}{\frac{1}{6}} \\
& \Rightarrow \frac{x-1}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{1}{3}}{1}
\end{aligned}\)
\(\therefore \quad\) The given line passes through \(\left(1,-\frac{1}{3}, \frac{1}{3}\right)\) and has direction ratios proportional to \(3,2,1\).
\(\therefore \quad\) Vector equation is \(\overline{\mathrm{r}}=\left(\hat{\mathrm{i}}-\frac{\hat{\mathrm{j}}}{3}+\frac{\hat{\mathrm{k}}}{3}\right)+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})\)
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