Q81 / 170 · Manipal (MET) 2019 · Mathematics
The projection of a line on a co-ordinate axes are $2,3,6$. Then the length of the line is
- Correct answerA$7$
- B$5$
- C$1$
- D$11$
Solution
Let $\mathbf{a}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}$
Here, $\left(a_1 \hat{\mathbf{i}}\right) \cdot \hat{\mathbf{i}}=a_1=2 ;\left(a_2 \hat{\mathbf{j}}\right) \cdot \hat{\mathbf{j}}=a_2=3$ and $\left(a_3 \hat{\mathbf{k}}\right) \cdot \hat{\mathbf{k}}=a_3=6$ $\therefore \mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ $\text { Now, required length } \begin{aligned} & =|a|=\sqrt{2^2+3^2+6^2} \\ & =\sqrt{4+9+36}=\sqrt{49}=7 \end{aligned}$
Here, $\left(a_1 \hat{\mathbf{i}}\right) \cdot \hat{\mathbf{i}}=a_1=2 ;\left(a_2 \hat{\mathbf{j}}\right) \cdot \hat{\mathbf{j}}=a_2=3$ and $\left(a_3 \hat{\mathbf{k}}\right) \cdot \hat{\mathbf{k}}=a_3=6$ $\therefore \mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ $\text { Now, required length } \begin{aligned} & =|a|=\sqrt{2^2+3^2+6^2} \\ & =\sqrt{4+9+36}=\sqrt{49}=7 \end{aligned}$
Hence, the correct answer is option: 'A'.