MHT CET · Maths · Vector Algebra
The vector equation of the line whose Cartesian equations are \(y=2\) and \(4 x-3 z+5=0\) is
- A \(\overline{\mathrm{r}}=(2 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})+\lambda(4 \hat{\mathrm{i}}-3 \hat{\mathrm{k}})\)
- B \(\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}-\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}})\)
- C \(\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}-\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}-4 \hat{\mathrm{k}})\)
- D \(\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}})\)
Answer & Solution
Correct Answer
(D) \(\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}})\)
Step-by-step Solution
Detailed explanation
We have \(4 x-3 z+5=0\) and \(y=2\)
\(
\begin{aligned}
& \therefore 4 x=3 z-5 \Rightarrow 4 x=3\left(z-\frac{5}{3}\right) \\
& \therefore \frac{4 x}{12}=\frac{3\left(z-\frac{5}{3}\right)}{12} \Rightarrow \frac{x}{3}=\frac{3\left(z-\frac{5}{3}\right)}{4}
\end{aligned}
\)
Thus line passes through point \(\left(0,2, \frac{5}{3}\right)\), and has direction ratios \(3,0,4\).
Hence required equation of line is \(\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}})\)
\(
\begin{aligned}
& \therefore 4 x=3 z-5 \Rightarrow 4 x=3\left(z-\frac{5}{3}\right) \\
& \therefore \frac{4 x}{12}=\frac{3\left(z-\frac{5}{3}\right)}{12} \Rightarrow \frac{x}{3}=\frac{3\left(z-\frac{5}{3}\right)}{4}
\end{aligned}
\)
Thus line passes through point \(\left(0,2, \frac{5}{3}\right)\), and has direction ratios \(3,0,4\).
Hence required equation of line is \(\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}})\)
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