Question

If \[ \vec{a}=(p,-2,5) \] and \[ \vec{b}=(1,q,-3) \] are collinear vectors then

• A. \(p=\dfrac{5}{3},\ q=\dfrac{6}{5}\)
• B. \(p=-\dfrac{5}{3},\ q=-\dfrac{6}{5}\)
• C. \(p=\dfrac{5}{3},\ q=-\dfrac{6}{5}\)
• D. \(p=-\dfrac{5}{3},\ q=\dfrac{6}{5}\)

Hint:

Two vectors are collinear if the ratios of their corresponding components are equal. \[ \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \]

Answer 1

The Correct Option is D

Step 1: Recall what collinear vectors mean.
Two vectors are collinear (parallel) if and only if one is a scalar multiple of the other. If $ \vec{a} = k \vec{b} $ for some scalar $ k $, then every pair of corresponding components satisfies the same ratio.
Step 2: Set up the scalar multiple equation.
Given $ \vec{a} = (p, -2, 5) $ and $ \vec{b} = (1, q, -3) $, we need $ \vec{a} = k \vec{b} $ for some constant $ k $. This gives $ p = k $, $ -2 = kq $, and $ 5 = -3k $.
Step 3: Solve for $ k $ from the third component.
From $ 5 = -3k $, we get \[ k = -\frac{5}{3} \] Therefore $ p = k = -\dfrac{5}{3} $.
Step 4: Solve for $ q $ from the second component.
We have $ -2 = kq = -\dfrac{5}{3} \cdot q $. Solving, \[ q = \frac{-2}{-5/3} = \frac{-2 \times 3}{-5} = \frac{6}{5} \]
Step 5: Verify consistency with all three components.
Check: $ \dfrac{p}{1} = -\dfrac{5}{3} $, $ \dfrac{-2}{q} = \dfrac{-2}{6/5} = -\dfrac{5}{3} $, $ \dfrac{5}{-3} = -\dfrac{5}{3} $. All ratios match, confirming collinearity.
Step 6: State the answer.
\[ \boxed{p = -\frac{5}{3},\quad q = \frac{6}{5}} \]