Three Dimensional Geometry - Online Test

Answer : Option D
Explaination / Solution:

Find the equation of the line in cartesian form that passes through the point (– 2, 4, – 5) and parallel to the line given by
x+33=y−45=z+86 
is given by:
x−x1l=y−y1m=z−z1n x1=−2,x2=4,x3=−5And l = 3 , m = 5 and n = 6 .
Therefore , 
x+23=y−45=z+56.

Answer : Option C
Explaination / Solution:

The coordinates of the origin are ( 0 , 0 ,0 ) , therefore , x1=0,x2=0,x3=0, and l = 5 , m = - 2 and n = 3 , 
therefore the equation in Cartesian form is given by:
x−x1l=y−y1m=z−z1n
i.e. x5=y−2=z3.

Answer : Option C
Explaination / Solution:

Here , a→=−3i^−2j^−5k^ and b→=0i^+0j^+11k^
Therefore , the vector equation is :
r→=a→+λb→ 
i.e.r⃗ =3i^−2j^−5k^+λ(11k^.).

Answer : Option B
Explaination / Solution:

Answer : Option D
Explaination / Solution:

Answer : Option C
Explaination / Solution:

Answer : Option D
Explaination / Solution:

On comparing the given equations with:r→=a1→+λb1→,and,r→=a2→+μb2→
, we get: 

Answer : Option C
Explaination / Solution:

The given equations can be reduced as: 

Answer : Option D
Explaination / Solution:

In the vector form, equation of a plane which is at a distance d from the origin, and n^ is the unit vector normal to the plane through the origin is given by : 

Answer : Option D
Explaination / Solution:

In Cartesian co – ordinate system Equation of a plane which is at a distance d from the origin and the direction cosines of the normal to the plane are l, m, n is given by : lx + my + nz = d.