Question

The d.r.'s of the line $AB$ are $(6,-2,9)$. If the line $AB$ makes angles $\alpha ,\beta$ with $OY,OZ$ respectively where $O=(0,0,0),$ then ${\sin }^2\alpha -{\sin }^2\beta =$

• A. $\frac{77}{121}$
• B. $\frac{-32}{121}$
• C. $\frac{40}{121}$
• D. $\frac{85}{121}$

Answer 1

The correct option is A $\frac{77}{121}$

$(a,b,c)=(6,-2,9)$
$\sqrt{a^2+b^2+c^2}=\sqrt{36+4+81}=\sqrt{121}$
$\cos \alpha =\frac{-2}{\sqrt{121}},\cos \beta =\frac{9}{\sqrt{121}}$
$\sin \alpha =\sqrt{1-(\cos \alpha {)}^2}=\sqrt{1-\frac{4}{121}}=\frac{\sqrt{117}}{11}$
$\sin \beta =\sqrt{1-(\cos \beta {)}^2}=\sqrt{1-\frac{81}{121}}=\frac{\sqrt{40}}{11}$

$\therefore {\sin }^2\alpha -{\sin }^2\beta =\frac{117}{121}-\frac{40}{121}=\frac{77}{121}$