| Statement-1: If \[\vec{a}=\vec{i}+\vec{j}\] and \[\vec{b}=\vec{j}-\vec{k}\] then \[(\vec{a}+\vec{b},\vec{a}-\vec{b})={{90}^{o}}\] |
| Statement-2 : Projection of \[\vec{a}+\vec{b}\] on \[\vec{a}-\vec{b}\]is zero. |
A) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
B) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
C) Statement-1 is true, Statement-2 is false.
D) Statement-1 is false, Statement-2 is true.
Correct Answer: B
Solution :
Statement 1 : \[\cos (\vec{a}+\vec{b},\vec{a}-\vec{b})=\frac{(\hat{i}-2\hat{j}-\hat{k}).(\hat{i}+\hat{k})}{\sqrt{6}\sqrt{2}}=\frac{1+0-1}{\sqrt{12}}=0;\]\[\Rightarrow \]\[(\vec{a}+\vec{b},\vec{a}-\vec{b})=90\] Statement 2 : \[\frac{(\vec{a}+\vec{b},\vec{a}-\vec{b})}{|\hat{a}-\hat{b}|}=\frac{0}{\sqrt{2}}=0\]
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