MHT CET · Maths · Three Dimensional Geometry
The vector equation of the line \(2 x+4=3 y+1=6 z-3\) is
- A \(\overline{\mathrm{r}}=\left(2 \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})\)
- B \(\overline{\mathrm{r}}=\left(-2 \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})\)
- C \(\overline{\mathrm{r}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})\)
- D \(\overline{\mathrm{r}}=(-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})\)
Answer & Solution
Correct Answer
(B) \(\overline{\mathrm{r}}=\left(-2 \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})\)
Step-by-step Solution
Detailed explanation
The equation of line is
\(
\begin{aligned}
& 2 x+4=3 y+1=6 z-3 \\
& \Rightarrow 2(x+2)=3\left(y+\frac{1}{3}\right)=6\left(z-\frac{1}{2}\right) \\
& \Rightarrow \frac{x+2}{\frac{1}{2}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-\frac{1}{2}}{\frac{1}{6}} \\
& \Rightarrow \frac{x+2}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{1}{2}}{1}
\end{aligned}
\)
\(\therefore \quad\) The given line passes through \(\left(-2,-\frac{1}{3}, \frac{1}{2}\right)\) and has direction ratios proportional to \(3,2,1\).
\(\therefore \quad\) Vector equation of the line is
\(
\overline{\mathrm{r}}=\left(-2 \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})
\)
\(
\begin{aligned}
& 2 x+4=3 y+1=6 z-3 \\
& \Rightarrow 2(x+2)=3\left(y+\frac{1}{3}\right)=6\left(z-\frac{1}{2}\right) \\
& \Rightarrow \frac{x+2}{\frac{1}{2}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-\frac{1}{2}}{\frac{1}{6}} \\
& \Rightarrow \frac{x+2}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{1}{2}}{1}
\end{aligned}
\)
\(\therefore \quad\) The given line passes through \(\left(-2,-\frac{1}{3}, \frac{1}{2}\right)\) and has direction ratios proportional to \(3,2,1\).
\(\therefore \quad\) Vector equation of the line is
\(
\overline{\mathrm{r}}=\left(-2 \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})
\)
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