Find the value of p such that the lines \(\frac{x+11}{4}=\frac{y+3}{-2}=\frac{z-3}{4} \,and \,\frac{x-3}{p}=\frac{y+12}{2}=\frac{z-3}{-12}\) are at right angles to each other.(a) p=11(b) p=12(c) p=13(d) p=4The question was asked during an interview.I'm obligated to ask this question of Three Dimensional Geometry in chapter Three Dimensional Geometry of Mathematics – Class 12

Find the value of p such that the lines \(\frac{x+11}{4}=\frac{y+3}{-2

1.	Find the value of p such that the lines \(\frac{x+11}{4}=\frac{y+3}{-2}=\frac{z-3}{4} \,and \,\frac{x-3}{p}=\frac{y+12}{2}=\frac{z-3}{-12}\) are at right angles to each other.(a) p=11(b) p=12(c) p=13(d) p=4The question was asked during an interview.I'm obligated to ask this question of Three Dimensional Geometry in chapter Three Dimensional Geometry of Mathematics – Class 12
Answer» The CORRECT choice is (c) p=13 To EXPLAIN: We know that, if TWO lines are perpendicular to each other then, \(a_1 a_2+b_1 b_2+c_1 c_2=0\) i.e.4(p)+(-2)2+4(-12)=0 4p-4-48=0 4p=52 p=\(\FRAC{52}{4}\)=13.

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