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3D Geometry question

2019 · 9 Apr · Shift 2 · Q38
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  5. /2019 · 9 Apr · Shift 2 · Q38

3D Geometry question

2019 · 9 Apr · Shift 2 · Q38

JEE MainMathematics3D GeometryMCQ+4 / −1
The vertices B and C of a Δ\DeltaΔ ABC lie on the line, x+23=y−10=z4{{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4}3x+2​=0y−1​=4z​ such that BC = 5 units. Then the area (in sq. units) of this triangle, given that the point A(1, –1, 2), is :
  1. A
    6
  2. B
    5175\sqrt {17}517​
  3. C
    34\sqrt {34}34​
  4. D
    2342\sqrt {34}234​
View written solutionFree

Correct answer: C

  1. Interpret the given line

The line is

x+23=y−10=z4.\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}.3x+2​=0y−1​=4z​.

Let the common parameter be ttt. Then

x+23=t,z4=t.\frac{x+2}{3}=t, \qquad \frac{z}{4}=t.3x+2​=t,4z​=t.

So,

x=3t−2,z=4t.x=3t-2, \qquad z=4t.x=3t−2,z=4t.

Also,

y−10=t\frac{y-1}{0}=t0y−1​=t

means y−1=0y-1=0y−1=0, hence

y=1.y=1.y=1.

So the line is

(x,y,z)=(−2,1,0)+t(3,0,4).(x,y,z)=(-2,1,0)+t(3,0,4).(x,y,z)=(−2,1,0)+t(3,0,4).

Thus, a point on the line is

P(−2,1,0)P(-2,1,0)P(−2,1,0)

and its direction vector is

d⃗=(3,0,4).\vec d=(3,0,4).d=(3,0,4).
  1. Use the fact that BBB and CCC lie on this line with BC=5BC=5BC=5

The magnitude of the direction vector is

∣d⃗∣=32+02+42=5.|\vec d|=\sqrt{3^2+0^2+4^2}=5.∣d∣=32+02+42​=5.

Hence, moving the parameter by 111 gives a distance of 555 units on the line. Since BC=5BC=5BC=5, the segment BCBCBC is a segment of length 555 along this line.

So for the area of triangle ABCABCABC,

Area=12×BC×(perpendicular distance of A from the line).\text{Area} = \frac12 \times BC \times (\text{perpendicular distance of }A\text{ from the line}).Area=21​×BC×(perpendicular distance of A from the line).

Thus we only need the perpendicular distance from A(1,−1,2)A(1,-1,2)A(1,−1,2) to the given line.

  1. Find the perpendicular distance from AAA to the line

Take point P(−2,1,0)P(-2,1,0)P(−2,1,0) on the line. Then

PA→=A−P=(1−(−2),−1−1,2−0)=(3,−2,2).\overrightarrow{PA}=A-P=(1-(-2),-1-1,2-0)=(3,-2,2).PA=A−P=(1−(−2),−1−1,2−0)=(3,−2,2).

Distance from point to line is

distance=∣PA→×d⃗∣∣d⃗∣.\text{distance} = \frac{|\overrightarrow{PA}\times \vec d|}{|\vec d|}.distance=∣d∣∣PA×d∣​.

Now,

PA→×d⃗=∣i^j^k^3−22304∣.\overrightarrow{PA}\times \vec d= \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & -2 & 2 \\ 3 & 0 & 4 \end{vmatrix}.PA×d=​i^33​j^​−20​k^24​​.

Compute:

=i^((−2)(4)−2(0))−j^(3⋅4−2⋅3)+k^(3⋅0−(−2)⋅3)=\hat i((-2)(4)-2(0)) - \hat j(3\cdot 4-2\cdot 3)+\hat k(3\cdot 0-(-2)\cdot 3)=i^((−2)(4)−2(0))−j^​(3⋅4−2⋅3)+k^(3⋅0−(−2)⋅3) =(−8)i^−(12−6)j^+6k^=(-8)\hat i -(12-6)\hat j +6\hat k=(−8)i^−(12−6)j^​+6k^ =(−8,−6,6).=(-8,-6,6).=(−8,−6,6).

Its magnitude is

(−8)2+(−6)2+62=64+36+36=136=234.\sqrt{(-8)^2+(-6)^2+6^2}= \sqrt{64+36+36}= \sqrt{136}=2\sqrt{34}.(−8)2+(−6)2+62​=64+36+36​=136​=234​.

Therefore,

distance=2345.\text{distance} = \frac{2\sqrt{34}}{5}.distance=5234​​.
  1. Compute the area of triangle ABCABCABC

Since

BC=5,BC=5,BC=5,

we get

Area=12×5×2345=34.\text{Area} = \frac12 \times 5 \times \frac{2\sqrt{34}}{5} =\sqrt{34}.Area=21​×5×5234​​=34​.
  1. Match with the options
34\boxed{\sqrt{34}}34​​

which corresponds to Option C.

  1. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.

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