Dimensional Geometry - 6 - Free MCQ Practice Test with solutions, Maths

Figure is self explanatory. From A OMP

A (r₁, θ) and B(r₂, θ₂) arc two given points through them passes the straight line L. Let P(r, θ) be an arbitrary point on L.

Then the points P, A and B are collinear.
=>the area of the triangle formed by these three points is zero

The other diagonal will always meet in a point.

x²- 7x + 6 - 0 => ( x - 1) ( x- 6) - 0
⇒ x= 1 and x = 6 are two opposite sides

=> y=4 and y = 10 are two opposite sides
Clearly the four vertices of the parallelogram are

The given straight line is yy' = 2a(x + x') ...(i)

the slope m of a straight line which ii perpendicular to (i) is given by

Now the required equation passes through (x',y') and posseses thes slope 
∴ Its equation will be

Let m' be the slope of the straight line which makes an angle a with the line 

Proof: Let m' be the slope of the straight line which makes an angle tan^-1 m with the straight line y = mx + c. Then

∴ The straight line with slope m₁ = 0 will be given by

The length of the perpendicular from (x₁, y₁ ,z₁) to the plane Ax+ By - Cz + D = 0  
is given by

∴ perpendicular distance d of the plane
6x-3y-2z -14=0
from the origin is given by

Note that an equation of first degree in x, y, z of the form
Ax + By + Cz + D = 0 ...(i)
represents a plane
d.r.’s of the normal to (i) are A,B,C.
∴ d.e.'s of the normal to (i) are

 Proof: The given planes are
ax + by + cz d = 0
and a'x + b'y + c'z + d' = 0 The d.r.’s of their normals are
a, b, c and a', b', c' repsectively.
∴ The d.c.’s of their normals are

Therefore if 0 is the required angle between the two planes (and hence between their normals), then

That is why statements (a), (b) and (c) are not correct.

Proof: The two planes are perpendicular if
θ = 90° or cos θ = 0
or aa' + bb' + cc' = 0
This is the condition of perpendicularity.
Remark : Condition of parallelism. Two planes are prallel if

The angle of inclination θ is given by

The condition that the four points (x₁, y₁, z₁), (x₂, y₂, z₂), (x₃, y₃, z₃) and (x4, y4, z4) are coplanar is that

Remark: The other three statements are same as he condition given above, (rows or columns in a determinant can be interchanged. Even number of changes do not change the value of the determinant. Odd number of changes make the value of the determinant as the negative of the original value)

Result: Two points P(x₁ , y₁, z₁) and Q (x₂ , y₂, z₂) lie: on the same side: or opposite sides of the: plane
Ax + By + Cz + D = 0 according as the quantities
Ax₁ + By₁ + Cz₁ + D and
Ax₂ + By₂ + Cz₂ + D are of same or opposite signs.

Proof: Substituting the coordinates of the given points P( 1,1, 1) and Q (-3, 0, 1) in the equation of plane, we get 

Since they are of opposite signs, therefore the points lie on the opposite sides of the plane. Also the perpendicular distances p and q of P and Q are given by

Thus the points are equidistant from the plane and on the opposite sides of it.

Length of perpendicular from origin