The vectors 3,i + j - 5,k and a,i + b,j - 15, | vedclass

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(D) Two vectors $\vec{u} = x_1\,i + y_1\,j + z_1\,k$ and $\vec{v} = x_2\,i + y_2\,j + z_2\,k$ are collinear if their components are proportional,i.e.,

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$a = 9, b = 3$

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(D) Two vectors $\vec{u} = x_1\,i + y_1\,j + z_1\,k$ and $\vec{v} = x_2\,i + y_2\,j + z_2\,k$ are collinear if their components are proportional,i.e.,$\frac{x_1}{x_2} = \frac{y_1}{y_2} = \frac{z_1}{z_2}$.Given vectors are $3\,i + j - 5\,k$ and $a\,i + b\,j - 15\,k$.Comparing the components,we get:$\frac{3}{a} = \frac{1}{b} = \frac{-5}{-15}$$\frac{3}{a} = \frac{1}{b} = \frac{1}{3}$From $\frac{3}{a} = \frac{1}{3}$,we get $a = 9$.From $\frac{1}{b} = \frac{1}{3}$,we get $b = 3$.Therefore,the correct option is $D$.

The vectors $3\,i + j - 5\,k$ and $a\,i + b\,j - 15\,k$ are collinear,if

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