KCET · Maths · Vector Algebra
The projection of \(\overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}\) on \(\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) is
- A \(\frac{8}{\sqrt{35}}\)
- B \(\frac{8}{\sqrt{39}}\)
- C \(\frac{8}{\sqrt{14}}\)
- D \(\sqrt{14}\)
Answer & Solution
Correct Answer
(C) \(\frac{8}{\sqrt{14}}\)
Step-by-step Solution
Detailed explanation
Given, \(\overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}\)
The projection of \(\overrightarrow{\mathbf{a}}\) on \(\overrightarrow{\mathbf{b}}=\frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{b}}|}\)
\(=\frac{(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{2^{2}+3^{2}+1^{2}}}\)
\[
=\frac{6-3+5}{\sqrt{14}}=\frac{8}{\sqrt{14}}
\]
The projection of \(\overrightarrow{\mathbf{a}}\) on \(\overrightarrow{\mathbf{b}}=\frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{b}}|}\)
\(=\frac{(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{2^{2}+3^{2}+1^{2}}}\)
\[
=\frac{6-3+5}{\sqrt{14}}=\frac{8}{\sqrt{14}}
\]
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