JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

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The line joining

A\left( {b\cos \mathrm{\alpha },b\sin \mathrm{\alpha }} \right)

and

B\left( {a\cos \mathrm{\beta },a\sin \mathrm{\beta }} \right)

is produced to the point

M\left( {x,y} \right)

so that

AM:MB = b:a

, then

x\cos \left( {\frac{{\mathrm{\alpha } + \mathrm{\beta }}}{2}} \right) + y\sin \left( {\frac{{\mathrm{\alpha } + \mathrm{\beta }}}{2}} \right)

View answer and explanation

Correct option: B

Option B (0) is correct because by using the section formula and trigonometric identities, the expression simplifies to 0.

More Questions from 3D GEOMETRY SYSTEM

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\left( { - 10,\;8} \right)

\left( { - 6,\;12} \right)

\left( {4,\; - 2} \right)

\left( { - 2,\;4} \right)

y z

(1,-2,4)

(3,-4

5)

p

2 x+y-3=0

3 x-y-2=0

x-p y+5=0

O\left( {0,0} \right),A\left( {a,0} \right)

B\left( {0,a} \right).

A(1,0,3)

B(4,7,1)

C(3,5,3)

\frac{x}{1}=\frac{y}{2}=\frac{z}{3}, \frac{x}{2}=\frac{y}{1}=\frac{z}{3}

\frac{x-1}{1}=\frac{2-y}{2}, z=3