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Plane

Equation of a Plane in Normal Form

• Two lines, , are coplanar, if and only if

• Two lines, ,

are coplanar, if

Solved Examples

Example 1:

Show that the lines are co-planar.

Solution:

The given lines can be rewritten as

Comparing with , we obtain

Hence, the given lines are co-planar.

Example 2:

For what value of p are the following two lines co-planar?

Solution:

Comparing the given lines with, we obtain

$\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right).\left(\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}\right)=\left(2\stackrel{^}{i}+5\stackrel{^}{j}-\stackrel{^}{k}\right).\left(-2p\stackrel{^}{i}+5\stackrel{^}{j}+p\stackrel{^}{k}\right)=-4p+25-p=-5p+25$

If these two lines are co-planar, then

Thus, the required value of p is 5.

• Two lines, , are coplanar, if and only if

• Two lines, ,

are coplanar, if

Solved Examples

Example 1:

Show that the lines are co-planar.

Solution:

The given lines can be rewritten as

Comparing with , we obtain

Hence, the given lines are co-planar.

Example 2:

For what value of p are the following two lines co-planar?

Solution:

Comparing the given lines with, we obtain

$\left(\stackrel{\to }{{a}_{2}}-\stackrel{\to }{{a}_{1}}\right).\left(\stackrel{\to }{{b}_{1}}×\stackrel{\to }{{b}_{2}}\right)=\left(2\stackrel{^}{i}+5\stackrel{^}{j}-\stackrel{^}{k}\right).\left(-2p\stackrel{^}{i}+5\stackrel{^}{j}+p\stackrel{^}{k}\right)=-4p+25-p=-5p+25$

If these two lines are co-planar, then

Thus, the required value of p is 5.

• If a plane is given whose perpendicular distance from origin is d, and is the unit normal vector along the direction of the normal to the plane from the origin, then:

• General vector equation of such a plane is given by

• Equation of the plane when direction cosines of the unit normal vector are given as l, m, and n is lx + my + nz = d

• Equation of the plane when the direction ratios of the unit normal vector are given as a, b, and c instead of direction cosines is:

Vector form− ; Cartesian form− ax + by + cz = d

• The normal drawn from origin to plane lx + my + nz = d will intersect the plane at (ld, md, nd).

• The normal drawn from origin to plane ax + bx + cz = d will intersect the plane at .

Solved Examples

Example 1:

What is the equation of the plane which is at the distance of units from the origin and its normal vector from the origin is?

Solution:

Normal vector,

Hence, the equation of plane is

Example 2:

Find the distance of plane 3x − 4y + 22 − 9 = 0 from the origin.

Solution:

3x − 4y + 2z = 9

Here, it can be noticed that

(3)2 + (− 4)2 + (2)2 = 29 1

Hence, equation is in the form of ax + by + cz = d

Dividing both sides by, we obtain

Distance of plane…

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