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MHT CET · Maths · Three Dimensional Geometry

The vector equation of the lien whose Cartesian equations are \(y\) \(=2\) and \(4 x-3 z+5=0\) is

  1. A \(\overline{\mathrm{r}}=(2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}-4 \hat{\mathrm{k}})\)
  2. B \(\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}})\)
  3. C \(\overline{\mathrm{r}}=(2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}})\)
  4. D \(\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}-4 \hat{\mathrm{k}})\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}})\)

Step-by-step Solution

Detailed explanation

We have lines \(y-2=0\) and \(4 x-3 z+5=0\)
\(\therefore 4 x=3 z-5=3\left[z-\left(\frac{5}{3}\right)\right] \)
\( \therefore \frac{4 x}{12}=\frac{3\left[z-\left(\frac{5}{3}\right)\right]}{12} \Rightarrow \frac{x}{3}=\frac{z-\left(\frac{5}{3}\right)}{4}, y=2\)
Thus line passes through the point \(\left(0,2, \frac{5}{3}\right)\) i.e. a point having position vector \(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\)
Also direction ratios of a line are \(3,0,4\) Hence required vector equation is
\(
\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}})
\)