Let be the co-ordinates of the foot of the perpendicular drawn from the point (5,4,2) on the line Then the length of the projection of the vector on the vector is [2026]

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Solution

Q.
Let $(α, β, γ)$ be the co-ordinates of the foot of the perpendicular drawn from the point (5,4,2) on the line $\vec{r} = (- \hat{i} + 3 \hat{j} + \hat{k}) + λ (2 \hat{i} + 3 \hat{j} - \hat{k}) .$ Then the length of the projection of the vector $α \hat{i} + β \hat{j} + γ \hat{k}$ on the vector $6 \hat{i} + 2 \hat{j} + 3 \hat{k}$ is [2026]

1 18/7

2 15/7

3 3

4 4

Ans.
(1)

$\vec{r} = (- \hat{i} + 3 \hat{j} + \hat{k}) + λ (2 \hat{i} + 3 \hat{j} - \hat{k})$

$\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1} = λ$

Any general point $P$ on the line is

$(2 λ - 1, 3 λ + 3, - λ + 1)$

Let the given point is A (5,4,2).

$\bar{A P} (2 λ - 6) \hat{i} + (3 λ - 1) \hat{j} + (- λ - 1) \hat{k}$

$∵$ $\bar{A P}$ $⊥^{r}$ $Line (L)$

$∴$ $\bar{A P} \cdot (2 \hat{i} + 3 \hat{j} - \hat{k}) = 0$

$2 (2 λ - 6) + 3 (3 λ - 1) - 1 (- λ - 1) = 0$

$\Rightarrow λ = 1$

$∴$ $α = 1, β = 6, γ = 0$

Let the vector $\bar{u} = α \hat{i} + β \hat{j} + γ \hat{k}$

$\bar{u} = \hat{i} + 6 \hat{j} + 0 \hat{k}$

$and$ $\bar{w} = 6 \hat{i} + 2 \hat{j} + 3 \hat{k}$

$So projection = \frac{| \bar{u} \cdot \bar{w} |}{| \bar{w} |} = \frac{18}{7}$

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