If the lines 1- x -3-7 y -14-2 p - z -3-2 and 7-7 x -3 p - y - x -1-6- z -5- and are at right angle- then the value of isA- 70 - 11B- 11 - 70C- 7 - 10D- 10 - 7

The correct option is A 70/11 a_1a_2+b_1b_2+c_1c_2=0 ⇒ 3 ( 3P/7 )+2P/7+2( 5)=0 ⇒ p=70/11 ...

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Byju's Answer

Standard XII

Mathematics

Equation of Conics in Complex Form

If the lines ...

Question

If the lines $\frac{1 - x}{3} = \frac{7 y - 14}{2 p} = \frac{z - 3}{2} a n d \frac{7 - 7 x}{3 p} = \frac{y - x}{1} = \frac{6 - z}{5} =$ and are at right angle, then the value of is

70/11

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11/70

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7/10

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10/7

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Solution

The correct option is A
70/11

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \Rightarrow - 3 (- \frac{3 P}{7}) + \frac{2 P}{7} + 2 (- 5) = 0 \Rightarrow p = \frac{70}{11}$

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Similar questions

Find the value of p so that the lines $\frac{1 - x}{3} = \frac{7 y - 14}{2 p} = \frac{z - 3}{2} a n d \frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}$ are at right angles

Q. Find the value of p so the line

\frac{1 - x}{3} = \frac{7 y - 14}{2 p} = \frac{z - 3}{2}

and

\frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}

are at right angles.

Q. Find the values of

p

so that the line

1 - x

\frac{1 - x}{3} = \frac{7 y - 14}{2 p} = \frac{z - 3}{2}

and

\frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}

are at right angles.

Q. The lines

\frac{1 - x}{3} = \frac{7 y - 14}{2 P} = \frac{z - 3}{2}

and

\frac{7 - 7 x}{3 P} = \frac{y - 5}{1} = \frac{6 - z}{5}

are perpendicular to each other. Find the value of

P

Q. Find the value of

λ

so that the lines.

\frac{1 - x}{3} = \frac{7 y - 14}{2 λ} = \frac{z - 3}{2}

and

\frac{7 - 7 x}{3 λ} = \frac{y - 5}{1} = \frac{6 - z}{5}

are at right angles

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If the lines 1−x3=7y−142p=z−32 and 7−7x3p=y−x1=6−z5= and are at right angle, then the value of is

The correct option is A 70/11 a1a2+b1b2+c1c2=0⇒−3(−3P7)+2P7+2(−5)=0⇒p=7011

If the lines $\frac{1 - x}{3} = \frac{7 y - 14}{2 p} = \frac{z - 3}{2} a n d \frac{7 - 7 x}{3 p} = \frac{y - x}{1} = \frac{6 - z}{5} =$ and are at right angle, then the value of is

The correct option is A
70/11

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \Rightarrow - 3 (- \frac{3 P}{7}) + \frac{2 P}{7} + 2 (- 5) = 0 \Rightarrow p = \frac{70}{11}$