Question

Find the equation of the line of intersection of planes 4x + 4y – 5z = 12 and 8x + 12y – 13z = 32 in the symmetric form.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A


We are given the equation of two planes. Two non- parallel planes intersect on a line or the intersection of two planes is a line. We know that they are non-parallel because the direction ratios of the normals of the planes are different. We can understand this by looking at the equations of the planes. We want to find the equation of that line in symmetrical form. Equation of a line in symmetrical form would look like :
$\frac{x-a}{u}=\frac{y-b}{v}=\frac{z-c}{w}$
Here, (a,b,c) are the coordinates of a point on the line and u,v,w are the direction ratios of the line.
Since the line would lie on both the planes, the vectors representing the direction ratios of the line and the normals of planes must be perpendicular.

Here, n1 and n2 are the normals and j is the vector in the direction of line.
So, the angle between j and n1 and j and n2 must be 90 degrees.
The two planes given to us are 4x + 4y – 5z = 12, 8x + 12y – 13z = 32 . The direction ratios of their normals are 4, 4, -5 and 8, 12, -13. Let l, m, n be the direction ratios of the line of intersection.
Direction ratios of the normals of the planes are 4,4, -5 and 8, 12, -13
Using the condition for perpendicularity of two line segments, we get
4l + 4m – 5n = 0
and 8l + 12m – 13n = 0
We want to find the ratios l,m and n. Since there are only two equations, we can assume some value for one of the three variables and find the other two values. Another way of solving this is using cross multiplication(explained at the end of solution) as
$\frac{1}{-52+60}=\frac{m}{-40+52}=\frac{n}{48-32}$
From this we get l:m: n = 8: 12: 16 = 2:3:4
So the direction ratios of the line are 2,3,4
Now we need one point on the line to write the equation of the line.
4x + 4y – 5z = 12 and 8x + 12y – 13z = 32 are the equations of the planes. We want to find a point which lies on both these planes or find some (x,y,z) such that it satisfies both the equations.
For that, put z =0 and solve the equation.
4x+4y = 12
8x+12y = 32
Solving these we get x = 1, y = 2
So, one of the points which lies on these two planes is (1,2 , 0)
We now have one point (1,2,0) and the direction ratios of the line (2,3,4)
Using this, we can write the equation of the line as
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z}{4}$
Cross Multiplication : If there are two equations of the form
ax+by+cz = 0 and
ex+fy+gz = 0, then x, y, z are related as
$\frac{x}{bg-cf}=\frac{y}{ec-ag}=\frac{z}{af-eb}$