JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

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Application +4 | -1

\mathrm{A}(1,-1,2), \mathrm{B}(5,7,-6), \mathrm{C}(3,4,-10)

and

\mathrm{D}(-1,-4,-2)

are the vertices of a quadrilateral

\mathrm{ABCD},

then its area is:

View answer and explanation

Correct option: A

12 \sqrt{29}

Option A (

12 \sqrt{29}

) is correct. The area of the quadrilateral is found by dividing it into two triangles and using the cross product to find the area of each triangle.

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