The value of k so that the lines x-1 -3 - y-2 2k - z-3 2- x-1 3k - y-5 1 - z-6 -5 are perpendicular to each other- is-

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Distance between Two Parallel Planes

The value of ...

Question

The value of $k$ so that the lines $\frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2}$ , $\frac{x - 1}{3 k} = \frac{y - 5}{1} = \frac{z - 6}{- 5}$ are perpendicular to each other, is:

- \frac{10}{7}

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- \frac{8}{7}

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- \frac{6}{7}

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None of these

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Solution

The correct option is A $- \frac{10}{7}$
Lines are perpendicular so the product of their corresponding normal vector is zero
$- 3 (3 k) + 2 k (1) + 2 (- 5) = 0$
$- 9 k + 2 k - 10 = 0$
$k = - \frac{10}{7}$

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Similar questions

Q. The value of

k

for which the lines

\frac{x + 1}{- 3} = \frac{y + 2}{2 k} = \frac{z - 3}{2} and \frac{x - 1}{3 k} = \frac{y + 5}{1} = \frac{z + 6}{7}

may be perpendicular is:

Q. If the line

\frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2}

and

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

are perpendicular to each other then

k =

Q. If the lines

\frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2}

and

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

are perpendicular, find the value of

k

If the lines $\frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2} a n d \frac{x - 1}{3 k} = \frac{y - 1}{1} = \frac{z - 6}{- 5}$ perpendicular, then find the value of k.

Q. Show that the lines

\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1}

and

\frac{x}{1} = \frac{y}{2} = \frac{z}{3}

are perpendicular to each other.

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The value of k so that the lines x−1−3=y−22k=z−32, x−13k=y−51=z−6−5 are perpendicular to each other, is:

The correct option is A −107Lines are perpendicular so the product of their corresponding normal vector is zero−3(3k)+2k(1)+2(−5)=0−9k+2k−10=0k=−107

The value of $k$ so that the lines $\frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2}$ , $\frac{x - 1}{3 k} = \frac{y - 5}{1} = \frac{z - 6}{- 5}$ are perpendicular to each other, is:

The correct option is A $- \frac{10}{7}$
Lines are perpendicular so the product of their corresponding normal vector is zero
$- 3 (3 k) + 2 k (1) + 2 (- 5) = 0$
$- 9 k + 2 k - 10 = 0$
$k = - \frac{10}{7}$