KCET · Maths · Three Dimensional Geometry
The equation of the line joining the points \((-3,4,11)\) and \((1,-2,7)\) is
- A \(\frac{x+3}{2}=\frac{y-4}{3}=\frac{z-11}{4}\)
- B \(\frac{x+3}{-2}=\frac{y-4}{3}=\frac{z-11}{2}\)
- C \(\frac{x+3}{-2}=\frac{y+4}{3}=\frac{z+11}{4}\)
- D \(\frac{x+3}{2}=\frac{y+4}{-3}=\frac{z+11}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{x+3}{-2}=\frac{y-4}{3}=\frac{z-11}{2}\)
Step-by-step Solution
Detailed explanation
Given, points are \((-3,4,11)\) and \((1,-2,7)\).
Let \(A(-3,4,11)\) and \(B(1,-2,7)\)
Direction ratios of \(A B=1-(-3),-2-4,7-11\)
\(=4,-6,-4\)
\(=-2,3,2\)
Now, we have one point \((-3,4,11)=\left(x_{1}, y_{1}, z_{1}\right)\) and direction ratios \((a, b, c)=-2,3,2\)
\(\because\) Equation of line :
\(\begin{aligned}
& \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} \\
\Rightarrow \quad \frac{x+3}{-2} &=\frac{y-4}{3}=\frac{z-11}{2}
\end{aligned}\)
Let \(A(-3,4,11)\) and \(B(1,-2,7)\)
Direction ratios of \(A B=1-(-3),-2-4,7-11\)
\(=4,-6,-4\)
\(=-2,3,2\)
Now, we have one point \((-3,4,11)=\left(x_{1}, y_{1}, z_{1}\right)\) and direction ratios \((a, b, c)=-2,3,2\)
\(\because\) Equation of line :
\(\begin{aligned}
& \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} \\
\Rightarrow \quad \frac{x+3}{-2} &=\frac{y-4}{3}=\frac{z-11}{2}
\end{aligned}\)
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