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Question: The planes 2x - 3y + z = 4 and x + 2y - 5z = 11 intersect in a line L. Then a vector parallel to L, ...

The planes 2x - 3y + z = 4 and x + 2y - 5z = 11 intersect in a line L. Then a vector parallel to L, is

A

13i^\hat{i}+11j^\hat{j}+7k^\hat{k}

B

13i^\hat{i}+11j^\hat{j}-7k^\hat{k}

C

13i^\hat{i}-11j^\hat{j}+7k^\hat{k}

D

i^\hat{i}+2j^\hat{j}-5k^\hat{k}

Answer

13i^\hat{i}+11j^\hat{j}+7k^\hat{k}

Explanation

Solution

The normals to the given planes are:

n1=(2,−3,1),n2=(1,2,−5)\mathbf{n}_1 = (2,-3,1), \quad \mathbf{n}_2 = (1,2,-5)

A direction vector for the line of intersection is the cross product:

d=n1×n2=(13,11,7)\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = (13, 11, 7)

Compute cross product of normals:

d=(2,−3,1)×(1,2,−5)=(13,11,7)\mathbf{d} = (2,-3,1) \times (1,2,-5) = (13, 11, 7)