Find the projection of vector \vec{b}=2\hat{i}+2\sqrt{2} \,\hat{j}-2\hat{k} on the vector \vec{a}=\hat{i}-\hat{j}-\sqrt{2} \,\hat{k}.

Category: QuestionsFind the projection of vector \vec{b}=2\hat{i}+2\sqrt{2} \,\hat{j}-2\hat{k} on the vector \vec{a}=\hat{i}-\hat{j}-\sqrt{2} \,\hat{k}.
Editor">Editor Staff asked 11 months ago

Find the projection of vector \vec{b}=2\hat{i}+2\sqrt{2} \,\hat{j}-2\hat{k} on the vector \vec{a}=\hat{i}-\hat{j}-\sqrt{2} \,\hat{k}.
 
(a) 2
 
(b) \sqrt{2}
 
(c) 1
 
(d) 2\sqrt{2}
 
I had been asked this question in an interview.
 
The doubt is from Product of Two Vectors-1 topic in division Vector Algebra of Mathematics – Class 12
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1 Answers
Editor">Editor Staff answered 11 months ago

Right answer is (b) \sqrt{2}
 
For explanation: The projection of vector \vec{b} on the vector \vec{b} is given by \frac{1}{|\vec{a}|} (\vec{a}.\vec{b})
 
|\vec{a}|=\sqrt{(1)^2+(-1)^2+(-\sqrt{2})^2}=\sqrt{1+1+2}=\sqrt{4}=2
 
Also, \vec{a}.\vec{b}=2(1)+2\sqrt{2} \,(-1)-2(-\sqrt{2})=2-2\sqrt{2}+2\sqrt{2}=2
 
Therefore, the projection of vector \hat{i}-\hat{j}-\sqrt{2} \,\hat{k} on the vector \vec{b}=2\hat{i}+2\sqrt{2}\hat{j}-2\hat{k}  is
 
\frac{1}{|\vec{a}|} (\vec{a}.\vec{b})=\frac{1}{2} (2)=1.