20.  Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:

              \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7} and \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}

Answers (1)

Given 

Two straight lines in 3D whose direction cosines (3,-16,7) and (3,8,-5)

Now the two vectors which are parallel to the two lines are

\vec a= 3\hat i-16\hat j+7\hat k and 

\vec b= 3\hat i+8\hat j-5\hat k

As we know, a vector perpendicular to both  vectors \vec a and \vec b is \vec a\times\vec b, so

\vec a\times\vec b=\begin{vmatrix} \hat i &\hat j &\hat k \\ 3& -16 &7 \\ 3&8 &-5 \end{vmatrix}=\hat i(80-56)-\hat j(-15-21)+\hat k(24+48)

\vec a\times\vec b=24\hat i+36\hat j+72\hat k

A vector parallel to this vector is 

\vec d=2\hat i+3\hat j+6\hat k

Now as we know the vector equation of the line which passes through point p and parallel to vector d is

L=\vec p+\lambda \vec d

Here in our question, give point p = (1,2,-4) which means position vector of this point is

\vec p = \hat i +2\hat j-4\hat k

So, the required line is 

L=\vec p+\lambda \vec d

L=\hat i+2\hat j-4\hat k +\lambda (2\hat i+3\hat j+6\hat k)

L=(2\lambda +1)\hat i+(2+3\lambda)\hat j+(6\lambda-4)\hat k

Preparation Products

Knockout KCET 2021

An exhaustive E-learning program for the complete preparation of KCET exam..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout KCET JEE Main 2021

It is an exhaustive preparation module made exclusively for cracking JEE & KCET.

₹ 27999/- ₹ 16999/-
Buy Now
Knockout NEET Sept 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
Buy Now
Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
Buy Now
Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Exams
Articles
Questions