The perpendicular distance of the point whose position vector is (1,3,5) from the line vecr= hati+2 hatj+3 hatk+λ ( hati+2 hatj+2 hatk) is equal to

Question

The perpendicular distance of the point whose position vector is (1,3,5) from the line vecr= hati+2 hatj+3 hatk+λ ( hati+2 hatj+2 hatk) is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

∴ P M=| vecv2| sin θ=√5 sin θ As, cos θ=( vecv1 ⋅ vecv2/| vecv1|| vecv2|)=(6/3 √5)=(2/√5) ⇒ sin θ=(1/√5) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/d7272f6b2af8ddd09477ca28efa22d4a-.png" /> ∴ P M=| vecv2| sin θ=√5((1/√5))=1

Q. The perpendicular distance of the point whose position vector is $(1, 3, 5)$ from the line $r = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ$ $(\hat{i} + 2 \hat{j} + 2 \hat{k})$ is equal to

$∴ PM = ∣ v_{2} ∣ sin θ = 5 sin θ$
As, $cos θ = \frac{v _{1} \cdot v _{2}}{∣ v _{1} ∣ ∣ v _{2} ∣} = \frac{6}{3 5} = \frac{2}{5}$
$\Rightarrow sin θ = \frac{1}{5}$

$∴ PM = ∣ v_{2} ∣ sin θ = 5 (\frac{1}{5}) = 1$

Q. The perpendicular distance of the point whose position vector is (1,3,5) from the line r=i^+2j^​+3k^+λ (i^+2j^​+2k^) is equal to

∴PM=∣v2​∣sinθ=5​sinθ As, cosθ=∣v1​∣∣v2​∣v1​⋅v2​​=35​6​=5​2​ ⇒sinθ=5​1​ ∴PM=∣v2​∣sinθ=5​(5​1​)=1

Q. The perpendicular distance of the point whose position vector is $(1, 3, 5)$ from the line $r = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ$ $(\hat{i} + 2 \hat{j} + 2 \hat{k})$ is equal to

$∴ PM = ∣ v_{2} ∣ sin θ = 5 sin θ$
As, $cos θ = \frac{v _{1} \cdot v _{2}}{∣ v _{1} ∣ ∣ v _{2} ∣} = \frac{6}{3 5} = \frac{2}{5}$
$\Rightarrow sin θ = \frac{1}{5}$

$∴ PM = ∣ v_{2} ∣ sin θ = 5 (\frac{1}{5}) = 1$