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3D Geometry question

2025 · 24 Jan · Shift 1 · Q35
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  5. /2025 · 24 Jan · Shift 1 · Q35

3D Geometry question

2025 · 24 Jan · Shift 1 · Q35

JEE MainMathematics3D GeometryMCQ+4 / −1
Let in a △ABC\triangle A B C△ABC, the length of the side ACA CAC be 6 , the vertex BBB be (1,2,3)(1,2,3)(1,2,3) and the vertices A,CA, CA,C lie on the line x−63=y−72=z−7−2\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}3x−6​=2y−7​=−2z−7​. Then the area (in sq. units) of △ABC\triangle A B C△ABC is:
  1. A
    42
  2. B
    17
  3. C
    56
  4. D
    21
View written solutionFree

Correct answer: D

  1. Interpret the given line

The line on which points AAA and CCC lie is

x−63=y−72=z−7−2=t.\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}=t.3x−6​=2y−7​=−2z−7​=t.

So its parametric form is

x=6+3t,y=7+2t,z=7−2t.x=6+3t,\quad y=7+2t,\quad z=7-2t.x=6+3t,y=7+2t,z=7−2t.

Hence a general point on the line is

P(t)=(6+3t, 7+2t, 7−2t).P(t)=(6+3t,\,7+2t,\,7-2t).P(t)=(6+3t,7+2t,7−2t).

The direction vector of the line is

d⃗=(3,2,−2).\vec d=(3,2,-2).d=(3,2,−2).

Its magnitude is

∣d⃗∣=32+22+(−2)2=17.|\vec d|=\sqrt{3^2+2^2+(-2)^2}=\sqrt{17}.∣d∣=32+22+(−2)2​=17​.
  1. Use the fact that AC=6AC=6AC=6

Since both AAA and CCC lie on this line, segment ACACAC lies along the line. So the base of the triangle is

AC=6.AC=6.AC=6.

Thus,

Area of △ABC=12×AC×(distance of B from the line).\text{Area of }\triangle ABC=\frac12\times AC\times (\text{distance of }B\text{ from the line}).Area of △ABC=21​×AC×(distance of B from the line).

So we need the perpendicular distance from B(1,2,3)B(1,2,3)B(1,2,3) to the given line.

  1. Find distance of point BBB from the line

Take a point on the line by putting t=0t=0t=0:

P0=(6,7,7).P_0=(6,7,7).P0​=(6,7,7).

Then

P0B→=B−P0=(1−6, 2−7, 3−7)=(−5,−5,−4).\overrightarrow{P_0B}=B-P_0=(1-6,\,2-7,\,3-7)=(-5,-5,-4).P0​B​=B−P0​=(1−6,2−7,3−7)=(−5,−5,−4).

Distance from point to line is

∣P0B→×d⃗∣∣d⃗∣.\frac{|\overrightarrow{P_0B}\times \vec d|}{|\vec d|}.∣d∣∣P0​B​×d∣​.

Now compute the cross product:

(−5,−5,−4)×(3,2,−2)=∣i^j^k^−5−5−432−2∣.(-5,-5,-4)\times(3,2,-2) =\begin{vmatrix} \hat i & \hat j & \hat k\\ -5 & -5 & -4\\ 3 & 2 & -2 \end{vmatrix}.(−5,−5,−4)×(3,2,−2)=​i^−53​j^​−52​k^−4−2​​.

Expanding,

=i^[(−5)(−2)−(−4)(2)]−j^[(−5)(−2)−(−4)(3)]+k^[(−5)(2)−(−5)(3)].=\hat i\big[(-5)(-2)-(-4)(2)\big] -\hat j\big[(-5)(-2)-(-4)(3)\big] +\hat k\big[(-5)(2)-(-5)(3)\big].=i^[(−5)(−2)−(−4)(2)]−j^​[(−5)(−2)−(−4)(3)]+k^[(−5)(2)−(−5)(3)].

So,

=i^(10+8)−j^(10+12)+k^(−10+15)=(18,−22,5).=\hat i(10+8)-\hat j(10+12)+\hat k(-10+15) =(18,-22,5).=i^(10+8)−j^​(10+12)+k^(−10+15)=(18,−22,5).

Its magnitude is

182+(−22)2+52=324+484+25=833=717.\sqrt{18^2+(-22)^2+5^2} =\sqrt{324+484+25} =\sqrt{833} =7\sqrt{17}.182+(−22)2+52​=324+484+25​=833​=717​.

Therefore the perpendicular distance is

71717=7.\frac{7\sqrt{17}}{\sqrt{17}}=7.17​717​​=7.
  1. Compute the area

Now,

Area=12×6×7=21.\text{Area}=\frac12\times 6\times 7=21.Area=21​×6×7=21.
  1. Check options

The area is

21,21,21,

which corresponds to Option D.

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