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Question: Let A be the point of intersection of the lines $L_1: \frac{x-1}{5}=\frac{y-2}{0}=\frac{z-3}{0}, L_2...

Let A be the point of intersection of the lines L1:x−15=y−20=z−30,L2:x−14=y−23=z−30L_1: \frac{x-1}{5}=\frac{y-2}{0}=\frac{z-3}{0}, L_2: \frac{x-1}{4}=\frac{y-2}{3}=\frac{z-3}{0}. Let B and C be points on L1L_1 and L2L_2 respectively such that AB=AC=10AB = AC = 10. Then the area of △ABC\triangle ABC is equal to (in square units)

A

15

B

30

C

48

D

22

Answer

30

Explanation

Solution

  1. Find the Intersection Point A: The equations of the lines are: L1:x−15=y−20=z−30L_1: \frac{x-1}{5}=\frac{y-2}{0}=\frac{z-3}{0} L2:x−14=y−23=z−30L_2: \frac{x-1}{4}=\frac{y-2}{3}=\frac{z-3}{0}

From L1L_1, we have y−2=0  ⟹  y=2y-2=0 \implies y=2 and z−3=0  ⟹  z=3z-3=0 \implies z=3. From L2L_2, we have z−3=0  ⟹  z=3z-3=0 \implies z=3.

Substitute y=2y=2 and z=3z=3 into L2L_2: x−14=2−23=3−30  ⟹  x−14=0=0\frac{x-1}{4}=\frac{2-2}{3}=\frac{3-3}{0} \implies \frac{x-1}{4} = 0 = 0. This gives x−1=0  ⟹  x=1x-1=0 \implies x=1. So, the intersection point A is (1,2,3)(1, 2, 3).

  1. Find the Position Vectors of B and C:

    • Point B on L1L_1: L1L_1 passes through A(1,2,3)A(1, 2, 3) with direction vector d1⃗=(5,0,0)\vec{d_1} = (5, 0, 0). The unit direction vector is d1^=(5,0,0)5=(1,0,0)\hat{d_1} = \frac{(5, 0, 0)}{5} = (1, 0, 0). Since AB=10AB = 10, AB⃗=10d1^=(10,0,0)\vec{AB} = 10 \hat{d_1} = (10, 0, 0). B=A+AB⃗=(1,2,3)+(10,0,0)=(11,2,3)B = A + \vec{AB} = (1, 2, 3) + (10, 0, 0) = (11, 2, 3).

    • Point C on L2L_2: L2L_2 passes through A(1,2,3)A(1, 2, 3) with direction vector d2⃗=(4,3,0)\vec{d_2} = (4, 3, 0). The unit direction vector is d2^=(4,3,0)42+32=(4,3,0)5=(45,35,0)\hat{d_2} = \frac{(4, 3, 0)}{\sqrt{4^2+3^2}} = \frac{(4, 3, 0)}{5} = (\frac{4}{5}, \frac{3}{5}, 0). Since AC=10AC = 10, AC⃗=10d2^=10(45,35,0)=(8,6,0)\vec{AC} = 10 \hat{d_2} = 10(\frac{4}{5}, \frac{3}{5}, 0) = (8, 6, 0). C=A+AC⃗=(1,2,3)+(8,6,0)=(9,8,3)C = A + \vec{AC} = (1, 2, 3) + (8, 6, 0) = (9, 8, 3).

  2. Calculate the Area of △ABC\triangle ABC: The area of a triangle formed by vectors AB⃗\vec{AB} and AC⃗\vec{AC} is given by 12∣AB⃗×AC⃗∣\frac{1}{2} |\vec{AB} \times \vec{AC}|. AB⃗=(10,0,0)\vec{AB} = (10, 0, 0) AC⃗=(8,6,0)\vec{AC} = (8, 6, 0)

    The cross product is:

    AB⃗×AC⃗=∣ijk1000860∣=i(0)−j(0)+k(10×6−0×8)=(0,0,60)\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 10 & 0 & 0 \\ 8 & 6 & 0 \end{vmatrix} = \mathbf{i}(0) - \mathbf{j}(0) + \mathbf{k}(10 \times 6 - 0 \times 8) = (0, 0, 60)

    The magnitude is ∣AB⃗×AC⃗∣=02+02+602=60|\vec{AB} \times \vec{AC}| = \sqrt{0^2 + 0^2 + 60^2} = 60.

    Area =12×60=30= \frac{1}{2} \times 60 = 30 square units.

Alternatively, using Area =12AB⋅ACsin⁡θ= \frac{1}{2} AB \cdot AC \sin \theta: AB=10AB = 10, AC=10AC = 10. AB⃗⋅AC⃗=(10)(8)+(0)(6)+(0)(0)=80\vec{AB} \cdot \vec{AC} = (10)(8) + (0)(6) + (0)(0) = 80. ∣AB⃗∣∣AC⃗∣cos⁡θ=10×10cos⁡θ=100cos⁡θ|\vec{AB}||\vec{AC}| \cos \theta = 10 \times 10 \cos \theta = 100 \cos \theta. 80=100cos⁡θ  ⟹  cos⁡θ=4580 = 100 \cos \theta \implies \cos \theta = \frac{4}{5}. sin⁡θ=1−(45)2=1−1625=925=35\sin \theta = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}. Area =12×10×10×35=50×35=30= \frac{1}{2} \times 10 \times 10 \times \frac{3}{5} = 50 \times \frac{3}{5} = 30.