Question

STA TEMENT-1 : For real numbers $l,m,n,\alpha ,\beta ,\gamma$ we have $l^2+m^2+n^2=2,{\alpha }^2+{\beta }^2+{\gamma }^2=18$ and ${(m\gamma -n\beta )}^2+(n\alpha -l\gamma {)}^2+(l\beta -m\alpha {)}^2=36$ Then $l\alpha +m\beta +n\gamma =0$.
STATEMENT-2 : If $\overline{u}$ and $\overline{v}$ are any two vectors, then $(\overline{u}.\overline{v}{)}^2={\overline{u}}^2{\overline{v}}^2-(\overline{u}\times \overline{v}{)}^2$

• A. SATEMENT -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

Answer 1

Answer: (A)

SATEMENT -1 is True, STATEMENT-2 is True;

STATEMENT-2 is a correct explanation for STATEMENT-1
Reason is true as it is a standard result
Assertion
Let $\overrightarrow{u}=l\hat{i}+m\hat{j}+n\hat{k}$ and $\overrightarrow{v}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$
Then $\overrightarrow{u}\cdot \overrightarrow{v}=l\alpha +m\beta +n\gamma$
Now ${\left|\overrightarrow{u}\right|}^2=l^2+m^2+n^2=2$ and ${\left|\overrightarrow{v}\right|}^2={\alpha }^2+{\beta }^2+{\gamma }^2=18$
Thus ${|\overrightarrow{u}\times \overrightarrow{v}|}^2={(m\gamma -n\beta )}^2+{(n\alpha -l\gamma )}^2+{(l\beta -m\alpha )}^2=36$
Therefore, ${(l\alpha +m\beta +n\gamma )}^2=2\times 18-36=0\Rightarrow l\alpha +m\beta +n\gamma =0$