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Question
What is the cosine of the angle which the vector \[\sqrt{2} \hat{i} + \hat{j} + \hat{k}\] makes with y-axis?
Solution
Given \[\sqrt{2} \hat{i} + \hat{j} + \hat{k}\].
Therefore , direction cosines are \[\frac{\sqrt{2}}{\sqrt{\left( \sqrt{2} \right)^2 + 1^2 + 1^2}} , \frac{1}{\sqrt{\left( \sqrt{2} \right)^2 + 1^2 + 1^2}} , \frac{1}{\sqrt{\left( \sqrt{2} \right)^2 + 1^2 + 1^2}}\] or \[\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\]
So, cosine angle with respect to y-axis is \[\frac{1}{2}\]
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