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