If a line having direction cosines 1, 3, 7 is perpendicular to a line with direction cosines 2, 3, K then the value of K will be -

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
We know the condition for perpendicularity of two lines i.e
$l_{1} . l_{2} + m_{1} . m_{2} + n_{1} . n_{2} = 0$
So, we’ll just substitute the corresponding values and find the value of K.
1.2 + 3.3 + 7.K = 0
$K = \frac{- 11}{7}$

Suggest Corrections

Similar questions

Q. If the direction cosines of a line are

- \frac{1}{2}, \frac{1}{3}

k then the value of k is

Q. Show that the three lines with direction cosines

\frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13} : \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{- 4}{13}, \frac{12}{13}

are mutually perpendicular

Show that the lines with direction cosines $\frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}$ are mutually perpendicular.

Q. Show that the three lines with direction cosines

\frac{12}{13} \frac{- 3}{13}, \frac{- 4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}

are mutually perpendicular.

Show that the three lines with direction cosines

are mutually perpendicular.

Join BYJU'S Learning Program

If a line having direction cosines 1, 3, 7 is perpendicular to a line with direction cosines 2, 3, K then the value of K will be -

The correct option is C We know the condition for perpendicularity of two lines i.e l1.l2+m1.m2+n1.n2=0 So, we’ll just substitute the corresponding values and find the value of K. 1.2 + 3.3 + 7.K = 0 K=−117

The correct option is C
We know the condition for perpendicularity of two lines i.e
$l_{1} . l_{2} + m_{1} . m_{2} + n_{1} . n_{2} = 0$
So, we’ll just substitute the corresponding values and find the value of K.
1.2 + 3.3 + 7.K = 0
$K = \frac{- 11}{7}$