MHT CET · Maths · Three Dimensional Geometry
The equation of line passing through the point \((1,2,3)\) and perpendicular to the lines \(\frac{x-2}{3}=\frac{y-1}{2}=\frac{z+1}{-2}\) and \(\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}\) is
- A \(\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(4 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})\)
- B \(\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-4 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})\)
- C \(\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})\)
- D \(\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})\)
Answer & Solution
Correct Answer
(C) \(\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})\)
Step-by-step Solution
Detailed explanation
Required line is perpendicular to the lines \(\frac{x-2}{3}=\frac{y-1}{2}=\frac{z+1}{-2}\) and \(\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}\)
\(\therefore \quad\) Required line is parallel to vector
\(\overline{\mathrm{b}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
3 & 2 & -2 \\
2 & -3 & 1
\end{array}\right|=-4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}}\)
\(\therefore \quad\) The equation of the required line is
\((\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-4 \hat{i}-7 \hat{j}-13 \hat{k})\)
\(\therefore \quad\) Required line is parallel to vector
\(\overline{\mathrm{b}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
3 & 2 & -2 \\
2 & -3 & 1
\end{array}\right|=-4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}}\)
\(\therefore \quad\) The equation of the required line is
\((\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-4 \hat{i}-7 \hat{j}-13 \hat{k})\)
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