If the lines (1-x/3)=(7 y-14/2 p)=(z-3/2) and (7-7 x/3 p)=(5-y/1)=(6-z/5) are orthogonal to each other, then the value of p is

Question

If the lines (1-x/3)=(7 y-14/2 p)=(z-3/2) and (7-7 x/3 p)=(5-y/1)=(6-z/5) are orthogonal to each other, then the value of p is (A) 1 (B) (70/11) (C) 7 (D) 5

Tardigrade Admin · Accepted Answer

The first lines is (x-1/-3)=(y-2/2 p / 7)=(z-3/2); parallel to V1=-3 hati+(2 p/7) hatj+2 hatk Second line is ((x-1)/-3 p / 7)=(y-5/-1)=(z-6/-5); parallel to v2=-(3 p/7) hati- hatj-5 hatk Now, -3 ⋅((-3 p /7))+((2 p /7)) ⋅(-1)+2(-5)=0 ∴ 7 p =70 ⇒ p =10

Q. If the lines $\frac{1 - x}{3} = \frac{7 y - 14}{2 p} = \frac{z - 3}{2}$ and $\frac{7 - 7 x}{3 p} = \frac{5 - y}{1} = \frac{6 - z}{5}$ are orthogonal to each other, then the value of $p$ is

1

$\frac{70}{11}$

7

5

Q. If the lines 31−x​=2p7y−14​=2z−3​ and 3p7−7x​=15−y​=56−z​ are orthogonal to each other, then the value of p is

1

1170​

7

5

The first lines is −3x−1​=2p/7y−2​=2z−3​; parallel to V1​=−3i^+72p​j^​+2k^ Second line is −3p/7(x−1)​=−1y−5​=−5z−6​; parallel to v2​=−73p​i^−j^​−5k^ Now, −3⋅(7−3p​)+(72p​)⋅(−1)+2(−5)=0 ∴7p=70⇒p=10

Q. If the lines $\frac{1 - x}{3} = \frac{7 y - 14}{2 p} = \frac{z - 3}{2}$ and $\frac{7 - 7 x}{3 p} = \frac{5 - y}{1} = \frac{6 - z}{5}$ are orthogonal to each other, then the value of $p$ is

$\frac{70}{11}$