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MHT CET · Maths · Vector Algebra

The Cartesian equation of a line is \(3 x+1=6 y-2=-z+1\), then its vector equation is

  1. A \(\overline{\mathrm{r}}=\left(\frac{-1}{3} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-6 \hat{\mathrm{k}})\)
  2. B \(\overline{\mathrm{r}}=(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-\hat{\mathrm{k}})\)
  3. C \(\overline{\mathrm{r}}=\left(\frac{-1}{3} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+6 \hat{\mathrm{k}})\)
  4. D \(\overline{\mathrm{r}}=\left(\frac{-1}{3} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+6 \hat{\mathrm{k}})\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\overline{\mathrm{r}}=\left(\frac{-1}{3} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+6 \hat{\mathrm{k}})\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & 3 x+1=6 y-2=-z+1 \\ & \therefore 3\left(x+\frac{1}{3}\right)=6\left(y-\frac{2}{6}\right)=-(z-1) \\ & \therefore \frac{\left(x+\frac{1}{3}\right)}{\left(\frac{1}{3}\right)}=\frac{\left(y-\frac{1}{3}\right)}{\left(\frac{1}{6}\right)}=\frac{(z-1)}{(-1)} \Rightarrow \text { d.r.s. are } 2,1,-6 \\ & \text { Thus } \bar{r}=\left(-\frac{1}{3} \hat{i}+\frac{1}{3} \hat{j}+\hat{k}\right)+\lambda(2 \hat{i}+\hat{j}-6 \hat{k})\end{aligned}\)