​ Consider a plane bar r .(6i – 3j – 2k) + 1 = 0 1. Find dc’s perpendicular to the plane. ​

Question

Consider a plane \(\bar r\).(6i – 3j – 2k) + 1 = 0

1. Find dc’s perpendicular to the plane. 
2. Find a vector of magnitude 14 units perpendicular to given plane. 
3. Find the equation of a line parallel to the above vector and passing through the point (1, 2, 1 ).

Answer 1

1. Given, 
\(\bar r\).(6i – 3j – 2k) + 1 = 0 ____(1)

Now, |6i – 3j – 2k| = \(\sqrt{36+9+4}\) = 7

∴ \(\frac{6}{7} i- \frac{3}{7}j - \frac{2}{7}k\) is a unit perpendicular to the plane (1)

⇒ the dc’s perpendicular to the plane (1) are \(\frac{6}{7}, -\frac{3}{7},-\frac{2}{7}.\)

2. We have, 
\(\frac{6}{7} i- \frac{3}{7}j - \frac{2}{7}k\) is a unit perpendicular to the Plane (1). Therefore, a vector of

magnitude 14 units perpendicular to the Plane (1) is 14(\(\frac{6}{7} i- \frac{3}{7}j - \frac{2}{7}k\))

⇒ 12i – 6j – 4k.

3. Equation of a line parallel to the vector 12i – 6j – 4k and passing through the point (1, 2, 1 )is given by