MHT CET · Maths · Three Dimensional Geometry
Let \(\quad \overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \quad \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \quad\) and
\(\overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}\) be three vectors. A vector in the
plane of \(\overrightarrow{\mathbf{b}}\) and \(\overrightarrow{\mathbf{c}}\) whose projection on \(\overrightarrow{\mathbf{a}}\) is of magnitude \(\sqrt{\frac{2}{3}}\), is
- A \(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}\)
- B \(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)
- C \(2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}\)
- D \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}}\)
Answer & Solution
Correct Answer
(A) \(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}\)
Step-by-step Solution
Detailed explanation
Any vector in the plane of \(\overrightarrow{\mathbf{b}}\) and \(\overrightarrow{\mathbf{c}}\) is \(\overrightarrow{\mathbf{r}}=m \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}\)
\(
=(m+1) \hat{\mathbf{i}}+(2 m+1) \hat{\mathbf{j}}+(-m-2) \hat{\mathbf{k}}
\)
Projection of \(\overrightarrow{\mathbf{r}}\) on \(\overrightarrow{\mathbf{a}}=\frac{\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|}=\left|\sqrt{\frac{2}{3}}\right|\)
\(
\begin{array}{l}
\therefore \frac{2(m+1)-(2 m+1)+(-m-2)}{\sqrt{6}}=\pm \sqrt{\frac{2}{3}} \\
\Rightarrow-m-1=\pm 2
\end{array}
\)
\(\Rightarrow\) \(m=-3\) and 1 Hence, \(\overrightarrow{\mathbf{r}}=-2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{r}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}\)
\(
=(m+1) \hat{\mathbf{i}}+(2 m+1) \hat{\mathbf{j}}+(-m-2) \hat{\mathbf{k}}
\)
Projection of \(\overrightarrow{\mathbf{r}}\) on \(\overrightarrow{\mathbf{a}}=\frac{\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|}=\left|\sqrt{\frac{2}{3}}\right|\)
\(
\begin{array}{l}
\therefore \frac{2(m+1)-(2 m+1)+(-m-2)}{\sqrt{6}}=\pm \sqrt{\frac{2}{3}} \\
\Rightarrow-m-1=\pm 2
\end{array}
\)
\(\Rightarrow\) \(m=-3\) and 1 Hence, \(\overrightarrow{\mathbf{r}}=-2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{r}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}\)
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