Let P be a point in the plane of the vectors such that P is equidistant from the lines AB and AC. If then the area of the triangle ABP is: [2026]

Question

Let P be a point in the plane of the vectors $\vec{A B} = 3 \hat{i} + \hat{j} - \hat{k} and \vec{A C} = \hat{i} - \hat{j} + 3 \hat{k}$ such that P is equidistant from the lines AB and AC. If $| \vec{A P} |$ $= \frac{\sqrt{5}}{2},$ then the area of the triangle ABP is: [2026]

Smart Achievers · Accepted Answer

(4)

$\cos 2 θ = \frac{3 - 1 - 3}{\sqrt{11} \cdot \sqrt{11}} = - \frac{1}{11}$

$1 - 2 \sin^{2} θ = - \frac{1}{11} \Rightarrow 2 \sin^{2} θ = \frac{12}{11} \Rightarrow \sin θ = \sqrt{\frac{6}{11}}$

$∴$ $Area (∆ A P B) = \frac{1}{2} \times \sqrt{11} \cdot \frac{\sqrt{5}}{2} \cdot \sqrt{\frac{6}{11}} = \frac{\sqrt{30}}{4}$

Solution

Q.
Let P be a point in the plane of the vectors $\vec{A B} = 3 \hat{i} + \hat{j} - \hat{k} and \vec{A C} = \hat{i} - \hat{j} + 3 \hat{k}$ such that P is equidistant from the lines AB and AC. If $| \vec{A P} |$ $= \frac{\sqrt{5}}{2},$ then the area of the triangle ABP is: [2026]

1 $\frac{\sqrt{26}}{4}$

2 2

3 $\frac{3}{2}$

4 $\frac{\sqrt{30}}{4}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let P be a point in the plane of the vectors AB→=3i^+j^-k^ and AC→=i^-j^+3k^ such that P is equidistant from the lines AB and AC. If |AP→|=52, then the area of the triangle ABP is: [2026]

1 264

2 2

3 32

4 304

Ans. (4) cos2θ=3-1-311·11=-111 1-2sin2θ=-111⇒2sin2θ=1211⇒sinθ=611 ∴ Area(∆APB)=12×11·52·611=304

Want Us to Email you About Special Offers & Updates?

Q.
Let P be a point in the plane of the vectors $\vec{A B} = 3 \hat{i} + \hat{j} - \hat{k} and \vec{A C} = \hat{i} - \hat{j} + 3 \hat{k}$ such that P is equidistant from the lines AB and AC. If $| \vec{A P} |$ $= \frac{\sqrt{5}}{2},$ then the area of the triangle ABP is: [2026]

1 $\frac{\sqrt{26}}{4}$

3 $\frac{3}{2}$

4 $\frac{\sqrt{30}}{4}$