MHT CET · Maths · Three Dimensional Geometry
The equation of the line, through \(A(1,2,3)\) and perpendicular to the vector \(2 \hat{i}+\hat{j}-\hat{k}\) and \(\hat{i}+3 \hat{j}+2 \hat{k}\), is
- A \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})\)
- B \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}-\hat{k})\)
- C \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}+\hat{j}-\hat{k})\)
- D \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\)
Answer & Solution
Correct Answer
(D) \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\)
Step-by-step Solution
Detailed explanation
Let \(\overline{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\),
\(\begin{aligned}
& \overline{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}} \\
& \overline{\mathrm{c}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}
\end{aligned}\)
\(\overline{\mathrm{b}} \times \overline{\mathrm{c}}\) is perpendicular to both \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\).
\(\therefore \quad \overline{\mathrm{b}} \times \overline{\mathrm{c}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
2 & 1 & -1 \\
1 & 3 & 2
\end{array}\right|=5 \hat{i}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
The direction ratios of the required line are \(5,-5,5\) i.e. \(1,-1,1\)
\(\therefore \quad\) The required equation of line is
\(\overline{\mathrm{r}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
\(\begin{aligned}
& \overline{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}} \\
& \overline{\mathrm{c}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}
\end{aligned}\)
\(\overline{\mathrm{b}} \times \overline{\mathrm{c}}\) is perpendicular to both \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\).
\(\therefore \quad \overline{\mathrm{b}} \times \overline{\mathrm{c}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
2 & 1 & -1 \\
1 & 3 & 2
\end{array}\right|=5 \hat{i}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
The direction ratios of the required line are \(5,-5,5\) i.e. \(1,-1,1\)
\(\therefore \quad\) The required equation of line is
\(\overline{\mathrm{r}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
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