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3D Geometry question

2023 · 29 Jan · Shift 1 · Q48
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  5. /2023 · 29 Jan · Shift 1 · Q48

3D Geometry question

2023 · 29 Jan · Shift 1 · Q48

JEE MainMathematics3D GeometryNumerical+4 / −1
Let the co-ordinates of one vertex of ΔABC\Delta ABCΔABC be A(0,2,α)A(0,2,\alpha)A(0,2,α) and the other two vertices lie on the line x+α5=y−12=z+43{{x + \alpha } \over 5} = {{y - 1} \over 2} = {{z + 4} \over 3}5x+α​=2y−1​=3z+4​. For α∈Z\alpha \in \mathbb{Z}α∈Z, if the area of ΔABC\Delta ABCΔABC is 21 sq. units and the line segment BCBCBC has length 2212\sqrt{21}221​ units, then α2\alpha^2α2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 9

  1. Interpret the geometry

The points BBB and CCC lie on the line

x+α5=y−12=z+43=t.\frac{x+\alpha}{5}=\frac{y-1}{2}=\frac{z+4}{3}=t.5x+α​=2y−1​=3z+4​=t.

So a parametric form of the line is

x=5t−α,y=2t+1,z=3t−4.x=5t-\alpha,\qquad y=2t+1,\qquad z=3t-4.x=5t−α,y=2t+1,z=3t−4.

Hence its direction vector is

d⃗=(5,2,3).\vec d=(5,2,3).d=(5,2,3).

Given:

  • A=(0,2,α)A=(0,2,\alpha)A=(0,2,α)
  • BC=221BC=2\sqrt{21}BC=221​
  • Area of △ABC=21\triangle ABC=21△ABC=21
  1. Use area formula with base BCBCBC

If we take BCBCBC as the base, then

Area=12×BC×h,\text{Area}=\frac12 \times BC \times h,Area=21​×BC×h,

where hhh is the perpendicular distance from AAA to the line containing B,CB,CB,C.

So,

21=12⋅221⋅h  ⟹  h=2121=21.21=\frac12\cdot 2\sqrt{21}\cdot h \implies h=\frac{21}{\sqrt{21}}=\sqrt{21}.21=21​⋅221​⋅h⟹h=21​21​=21​.

Thus, the distance from AAA to the given line is 21\sqrt{21}21​.

  1. Distance from a point to a line in 3D

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

P=(−α,1,−4).P=(-\alpha,1,-4).P=(−α,1,−4).

Then

PA→=A−P=(0+α,2−1,α+4)=(α,1,α+4).\overrightarrow{PA}=A-P=(0+\alpha,2-1,\alpha+4)=(\alpha,1,\alpha+4).PA=A−P=(0+α,2−1,α+4)=(α,1,α+4).

Distance from point AAA to the line is

∥PA→×d⃗∥∥d⃗∥.\frac{\|\overrightarrow{PA}\times \vec d\|}{\|\vec d\|}.∥d∥∥PA×d∥​.

Here,

d⃗=(5,2,3),∥d⃗∥=25+4+9=38.\vec d=(5,2,3),\qquad \|\vec d\|=\sqrt{25+4+9}=\sqrt{38}.d=(5,2,3),∥d∥=25+4+9​=38​.

Now compute the cross product:

PA→×d⃗=∣i^j^k^α1α+4523∣.\overrightarrow{PA}\times \vec d= \begin{vmatrix} \hat i & \hat j & \hat k\\ \alpha & 1 & \alpha+4\\ 5 & 2 & 3 \end{vmatrix}.PA×d=​i^α5​j^​12​k^α+43​​.

Expanding,

=i^ (1⋅3−(α+4)⋅2)−j^ (α⋅3−(α+4)⋅5)+k^ (α⋅2−1⋅5).=\hat i\,(1\cdot 3-(\alpha+4)\cdot 2) -\hat j\,(\alpha\cdot 3-(\alpha+4)\cdot 5) +\hat k\,(\alpha\cdot 2-1\cdot 5).=i^(1⋅3−(α+4)⋅2)−j^​(α⋅3−(α+4)⋅5)+k^(α⋅2−1⋅5).

So,

PA→×d⃗=(−2α−5,  2α+20,  2α−5).\overrightarrow{PA}\times \vec d= (-2\alpha-5,\; 2\alpha+20,\; 2\alpha-5).PA×d=(−2α−5,2α+20,2α−5).

Therefore,

∥PA→×d⃗∥2=(−2α−5)2+(2α+20)2+(2α−5)2.\|\overrightarrow{PA}\times \vec d\|^2 =(-2\alpha-5)^2+(2\alpha+20)^2+(2\alpha-5)^2.∥PA×d∥2=(−2α−5)2+(2α+20)2+(2α−5)2.

Compute:

(−2α−5)2=4α2+20α+25,(-2\alpha-5)^2=4\alpha^2+20\alpha+25,(−2α−5)2=4α2+20α+25, (2α+20)2=4α2+80α+400,(2\alpha+20)^2=4\alpha^2+80\alpha+400,(2α+20)2=4α2+80α+400, (2α−5)2=4α2−20α+25.(2\alpha-5)^2=4\alpha^2-20\alpha+25.(2α−5)2=4α2−20α+25.

Adding,

∥PA→×d⃗∥2=12α2+80α+450.\|\overrightarrow{PA}\times \vec d\|^2=12\alpha^2+80\alpha+450.∥PA×d∥2=12α2+80α+450.
  1. Set the distance equal to 21\sqrt{21}21​

Since distance is 21\sqrt{21}21​,

12α2+80α+45038=21.\frac{\sqrt{12\alpha^2+80\alpha+450}}{\sqrt{38}}=\sqrt{21}.38​12α2+80α+450​​=21​.

Squaring both sides:

12α2+80α+45038=21.\frac{12\alpha^2+80\alpha+450}{38}=21.3812α2+80α+450​=21.

Hence,

12α2+80α+450=798.12\alpha^2+80\alpha+450=798.12α2+80α+450=798.

So,

12α2+80α−348=0.12\alpha^2+80\alpha-348=0.12α2+80α−348=0.

Divide by 444:

3α2+20α−87=0.3\alpha^2+20\alpha-87=0.3α2+20α−87=0.
  1. Solve for integer α\alphaα
3α2+20α−87=0.3\alpha^2+20\alpha-87=0.3α2+20α−87=0.

Discriminant:

Δ=202−4⋅3⋅(−87)=400+1044=1444=382.\Delta=20^2-4\cdot 3\cdot (-87)=400+1044=1444=38^2.Δ=202−4⋅3⋅(−87)=400+1044=1444=382.

Thus,

α=−20±386.\alpha=\frac{-20\pm 38}{6}.α=6−20±38​.

So,

α=3orα=−293.\alpha=3 \quad \text{or} \quad \alpha=-\frac{29}{3}.α=3orα=−329​.

Since α∈Z\alpha\in \mathbb Zα∈Z, we get

α=3.\alpha=3.α=3.

Therefore,

α2=9.\alpha^2=9.α2=9.
  1. Comparison with stored answer

Derived answer: 999. Stored correct answer: 999. They agree.

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