Let = 4 i ^ + 3 j ^ + 5 k ^ and = i ^ + 2 j ^ - 4 k ^ . Let 1 be parallel to and 2 be perpendicular to . If = 1 + 2 , then the value of 5 2 · ( i ^ + j ^ + k ^ ) is [2023]

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Solution

Q.
Let $\vec{α} = 4 \hat{i} + 3 \hat{j} + 5 \hat{k}$ and $\vec{β} = \hat{i} + 2 \hat{j} - 4 \hat{k}$ . Let ${\vec{β}}_{1}$ be parallel to $\vec{α}$ and ${\vec{β}}_{2}$ be perpendicular to $\vec{α}$ . If $\vec{β} = {\vec{β}}_{1} + {\vec{β}}_{2}$ , then the value of $5 {\vec{β}}_{2} \cdot (\hat{i} + \hat{j} + \hat{k})$ is [2023]

1 9

2 11

3 7

4 6

Ans.
(3)

$Given \vec{α} = 4 \hat{i} + 3 \hat{j} + 5 \hat{k} and \vec{β} = \hat{i} + 2 \hat{j} - 4 \hat{k}$

$As, {\vec{β}}_{1} ∥ \vec{α} \Rightarrow {\vec{β}}_{1} = λ \vec{α}$

$\Rightarrow {\vec{β}}_{1} = λ (4 \hat{i} + 3 \hat{j} + 5 \hat{k})$ ...(i)

${\vec{β}}_{2} ⊥ \vec{α} \Rightarrow {\vec{β}}_{2} \cdot \vec{α} = 0$

$Let {\vec{β}}_{2} = x \hat{i} + y \hat{j} + z \hat{k}$

$\Rightarrow 4 x + 3 y + 5 z = 0$ ...(ii) $[{\vec{β}}_{2} \cdot \vec{α} = 0]$

$Now, \vec{β} = {\vec{β}}_{1} + {\vec{β}}_{2}$

$\Rightarrow \hat{i} + 2 \hat{j} - 4 \hat{k} = (4 λ + x) \hat{i} + (3 λ + y) \hat{j} + (5 λ + z) \hat{k}$

$On comparing, we get$

$4 λ + x = 1 \Rightarrow x = 1 - 4 λ$ ...(iii)

$3 λ + y = 2 \Rightarrow y = 2 - 3 λ$ ...(iv)

$5 λ + z = - 4 \Rightarrow z = - 4 - 5 λ$ ...(v)

$Putting the value of x, y, z in (ii), we get$

$4 (1 - 4 λ) + 3 (2 - 3 λ) + 5 (- 4 - 5 λ) = 0$

$\Rightarrow 4 - 16 λ + 6 - 9 λ - 20 - 25 λ = 0$

$\Rightarrow - 10 - 50 λ = 0 \Rightarrow - 50 λ = 10 \Rightarrow λ = - \frac{1}{5}$

$Put the value of λ in (iii), (iv), (v), we get$

$∴$ $x = \frac{9}{5}, y = \frac{13}{5}, z = - 3$ $∴$ ${\vec{β}}_{2} = \frac{9}{5} \hat{i} + \frac{13}{5} \hat{j} - 3 \hat{k}$

$∴$ $5 {\vec{β}}_{2} \cdot (\hat{i} + \hat{j} + \hat{k}) = 9 + 13 - 15 = 7$

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