JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

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Application +4 | -1

Let the point, on the line passing through the points

{P}(1,-2,3)

and

Q(5,-4,7)

, farther from the origin and at a distance of 9 units from the point

P

, be

(\alpha, \beta, \gamma)

. Then

\alpha^2+\beta^2+\gamma^2

is equal to:

View answer and explanation

Correct option: A

155

Option D (

\frac{\pi}{32}

) is correct because, after solving the differential equation and applying the initial condition, y(2) equals

\frac{\pi}{32}

More Questions from 3D GEOMETRY SYSTEM

Practice All Chapter MCQs →

\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