JEE Main 2025 Jan 24 Shift 1, Mathematics Q8: Scalar (or Dot) Product of Two Vectors

JEE Main2025Jan 24, Shift 1Mathematics

Let in a $△ A B C$ , the length of the side $A C$ be 6, the vertex $B$ be $(1, 2, 3)$ and the vertices $A, C$ lie on the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Then the area (in sq. units) of $△ A B C$ is

A
$42$
B
$21$
C
$56$
D
$17$

Solution

The required height of the triangle from $B$ onto $A C$ is the perpendicular distance from $B (1, 2, 3)$ to the line containing $A C$ .

Let the foot of perpendicular be $M = (6 + 3 t, 7 + 2 t, 7 - 2 t)$ .

$B M = ⟨ 5 + 3 t, 5 + 2 t, 4 - 2 t ⟩$ . The direction of $A C$ is $d = ⟨ 3, 2, - 2 ⟩$ .

Set $B M \cdot d = 0$ :

$3 (5 + 3 t) + 2 (5 + 2 t) - 2 (4 - 2 t) = 15 + 9 t + 10 + 4 t - 8 + 4 t = 17 + 17 t = 0$

$t = - 1$ , giving $B M = ⟨ 2, 3, 6 ⟩$ and $∣ B M ∣ = 4 + 9 + 36 = 7$ .

Area $= \frac{1}{2} \cdot ∣ A C ∣ \cdot ∣ B M ∣ = \frac{1}{2} \cdot 6 \cdot 7 = 21$ .

$∴$ Option (2).

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JEE Main 2025 Jan 24 Shift 1, Mathematics Q8: Scalar (or Dot) Product of Two Vectors

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