MHT CET · Maths · Vector Algebra
The point \(\mathrm{P}\) lies on the line \(\mathrm{A} \mathrm{B}\), where \(\mathrm{A} \equiv(2,4,5)\) and \(\mathrm{B} \equiv(1,2,3)\). If \(\mathrm{z}\) co-ordinate
of point \(\mathrm{P}\) is 3, the its y co-ordinate is
- A 2
- B \(-2\)
- C \(-3\)
- D 3
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
(C)
Equation of line passing through \(\mathrm{A}\) and \(\mathrm{B}\) is
\(\frac{x-2}{1-2}=\frac{y-4}{2-4}=\frac{z-5}{3-5}=k \quad \text {...(say) } \Rightarrow \frac{x-2}{-1}=\frac{y-4}{-2}=\frac{z-5}{-2}=k\)
Hence coordinates of any point onthis line are
\((-k+2,-2 k+4,-2 k+5)\)
As per condition given, we have
\(-2 k+5=3 \Rightarrow k=1\)
Hence \(y\) coordinate \(=-2+4=2\)
Equation of line passing through \(\mathrm{A}\) and \(\mathrm{B}\) is
\(\frac{x-2}{1-2}=\frac{y-4}{2-4}=\frac{z-5}{3-5}=k \quad \text {...(say) } \Rightarrow \frac{x-2}{-1}=\frac{y-4}{-2}=\frac{z-5}{-2}=k\)
Hence coordinates of any point onthis line are
\((-k+2,-2 k+4,-2 k+5)\)
As per condition given, we have
\(-2 k+5=3 \Rightarrow k=1\)
Hence \(y\) coordinate \(=-2+4=2\)
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