If the points with position vectors $60\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}},40\widehat{\dot{\mathbf{i}}} - 8\widehat{\dot{\mathbf{j}}}$ and $a\widehat{\dot{\mathbf{i}}} - 52\widehat{\dot{\mathbf{j}}}$ are collinear, then $a$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
Let $P\left( 60\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} ight),\ Q\left( 40\widehat{\dot{\mathbf{i}}} - 8\widehat{\dot{\mathbf{j}}} ight)$and $R(a\widehat{\dot{\mathbf{i}}} - 52\widehat{\dot{\mathbf{j}}})$ be the collinear points. Then $\overrightarrow{\mathbf{PQ}} = \lambda\overrightarrow{\mathbf{QR}}\$
for some scalar λ
$\Longrightarrow \left( - 20\widehat{\dot{\mathbf{i}}} - 11\widehat{\dot{\mathbf{j}}} ight) = \lambda\left\lbrack (a - 40)\widehat{\dot{\mathbf{i}}} - 44\widehat{\dot{\mathbf{j}}} ightbrack$
$\Longrightarrow \ \lambda(a - 40) = - 20, - 44\lambda = - 11$
$\Longrightarrow \ \lambda(a - 40) = - 20,\ \lambda = \frac{1}{4}$
$	herefore a - 40 = - 20 	imes 4 \Longrightarrow a = - 40$

Q. If the points with position vectors $60\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}},40\widehat{\dot{\mathbf{i}}} - 8\widehat{\dot{\mathbf{j}}}$ and $a\widehat{\dot{\mathbf{i}}} - 52\widehat{\dot{\mathbf{j}}}$ are collinear, then $a$ is equal to

Explanation

Reviewed by Dr. Abubacker

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