KCET · Maths · Vector Algebra
The component of \(\mathbf{i}\) in the direction of the vector \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) is
- A \(6\)
- B \(6 \sqrt{6}\)
- C \(\frac{\sqrt{6}}{6}\)
- D \(\sqrt{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sqrt{6}}{6}\)
Step-by-step Solution
Detailed explanation
We know that component of a in the direction
of \(\mathbf{b}\) is
\(\frac{|\mathbf{a} \cdot \mathbf{b}|}{|\mathbf{b}|^2}\)
Let \(\mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) and \(\mathbf{a}=\hat{\mathbf{i}}\)
Then, we have \(|\mathbf{a} \cdot \mathbf{b}|=1\)
and \(|\mathbf{b}|=\sqrt{6}\)
Now, the component of \(i\) in the direction \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) is \(\frac{1}{\sqrt{6}}=\frac{\sqrt{6}}{6}\)
of \(\mathbf{b}\) is
\(\frac{|\mathbf{a} \cdot \mathbf{b}|}{|\mathbf{b}|^2}\)
Let \(\mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) and \(\mathbf{a}=\hat{\mathbf{i}}\)
Then, we have \(|\mathbf{a} \cdot \mathbf{b}|=1\)
and \(|\mathbf{b}|=\sqrt{6}\)
Now, the component of \(i\) in the direction \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) is \(\frac{1}{\sqrt{6}}=\frac{\sqrt{6}}{6}\)
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