TS EAMCET · Maths · Vector Algebra
The position vectors of three points \(A, B\) and \(C\) \((1,3, x),(3,5,8)\) and \((y,-1,-6)\) respectively. If \(A, B\) and \(C\) are collinear, then \((x, y)=\)
- A \(\left(\frac{2}{3},-3\right)\)
- B \(\left(\frac{10}{3}, 3\right)\)
- C \(\left(\frac{10}{3},-3\right)\)
- D \(\left(-3, \frac{10}{3}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{10}{3},-3\right)\)
Step-by-step Solution
Detailed explanation
We have, \(\begin{aligned} & A(1,3, x), B(3,5,8), C(y,-1,-6) \\ & \therefore \mathbf{A B}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+(8-x) \hat{\mathbf{k}} \\ & \mathbf{A C}=(y-1) \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+(-6-x) \hat{\mathbf{k}} \end{aligned}\) Since \(A, B, C\) are…
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