Question

Find the vector and cartesian equation of the plane passing through the points A( 2,3,1) , B(4,-5,3) and parallel to X-axis.
Please help me this is very urgent!

Answer 1

Answer:

let the required equation be by+cz+d=0-------1

since it passe through the point A(2,3,1)

Answer 2

The vector equation is 2j + 8k and plane equation is y + 4z = 7 which passes through A( 2,3,1) , B(4,-5,3) and parallel to X-axis.

Given:

Plane passes through points A( 2,3,1) and B(4,-5,3) and parallel to X-axis.

To find:

Vector and cartesian equation of the plane.

Solution:

Vector AB present on plane = (4i - 5j + 3k) - (2i + 3j + k) = 2i - 8j + 2k

Direction ratio of X axis = (1,0,0) = i + 0j +0k

Normal of plane = $\left[\begin{array}{ccc}i&j&k\\2&-8&2\\1&0&0\end{array}\right]$  = i(0) - j(0 - 2) + k(0 + 8)  = 2j + 8k

So vector equation of plane = n = 2j + 8k

We know that equation of plane = (r - a) n = 0

a = (2,3,1)

Vector form of a = 2i + 3j + k

Now, r . n = a . n (dot product)

⇒ r . (2j + 8k) = (2i + 3j + k) . (2j + 8k)

⇒ (xi + yj + zk) . (2j + 8k) = 2(0) + 3(2) + 1(8)

⇒ 2y + 8z = 14

⇒ y + 4z = 7

Therefore vector equation is 2j + 8k and plane equation is y + 4z = 7.

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