SATHEE: Vectors 2 Question 4

Vectors 2 Question 4

4. Let $\vec{α} = 3 \hat{i} + \hat{j}$ and $\vec{β} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ . If $\vec{β} = {\vec{β}}_{1} - {\vec{β}}_{2}$ , where ${\vec{β}}_{1}$ is parallel to $\vec{α}$ and ${\vec{β}}_{2}$ is perpendicular to $\vec{α}$ , then ${\vec{β}}_{1} \times {\vec{β}}_{2}$ is equal to

(a) $\frac{1}{2} (3 \hat{i} - 9 \hat{j} + 5 \hat{k})$

(b) $\frac{1}{2} (- 3 \hat{i} + 9 \hat{j} + 5 \hat{k})$

(d) $3 \hat{i} - 9 \hat{j} - 5 \hat{k}$

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Answer:

Correct Answer: 4. (c)

Solution:

Given vectors $\vec{α} = 3 \hat{i} + \hat{j}$ and $\vec{β} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ and $\vec{β} = {\vec{β}}_{1} - {\vec{β}}_{2}$ such that ${\vec{β}}_{1}$ is parallel to $\vec{α}$ and ${\vec{β}}_{2}$ is perpendicular to $α$

$\begin{aligned} So, {\vec{β}}_{1} = λ α = λ (3 \hat{i} + \hat{j}) \\ Now, {\vec{β}}_{2} = {\vec{β}}_{1} - \vec{β} = λ (3 \hat{i} + \hat{j}) - (2 \hat{i} - \hat{j} + 3 \hat{k}) \\ = (3 λ - 2) \hat{i} + (λ + 1) \hat{j} - 3 \hat{k} \end{aligned}$

$∵ {\vec{β}}_{2}$ is perpendicular to $α$ , so ${\vec{β}}_{2} \cdot α = 0$

[since if non-zero vectors $a$ and $b$ are perpendicular to each other, then $a \cdot b = 0]$

$\begin{array}{lr} ∴ & (3 λ - 2) (3) + (λ + 1) (1) = 0 \\ \Rightarrow & 9 λ - 6 + λ + 1 = 0 \\ \Rightarrow & 10 λ = 5 \Rightarrow λ = \frac{1}{2} \end{array}$

So, ${\vec{β}}_{1} = \frac{3}{2} \hat{i} + \frac{1}{2} \hat{j}$

and ${\vec{β}}_{2} = (\frac{3}{2} - 2) \hat{i} + (\frac{1}{2} + 1) \hat{j} - 3 \hat{k} = - \frac{1}{2} \hat{i} + \frac{3}{2} \hat{j} - 3 \hat{k}$

$\begin{aligned} ∴ {\vec{β}}_{1} \times {\vec{β}}_{2} & = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{3}{2} & \frac{1}{2} & 0 \\ \frac{- 1}{2} & \frac{3}{2} & - 3 \end{array} | = \hat{i} (- \frac{3}{2} - 0) - \hat{j} (- \frac{9}{2} - 0) + \hat{k} (\frac{9}{4} + \frac{1}{4}) \\ = - \frac{3}{2} \hat{i} + \frac{9}{2} \hat{j} + \frac{5}{2} \hat{k} = \frac{1}{2} (- 3 \hat{i} + 9 \hat{j} + 5 \hat{k}) \end{aligned}$

Vectors 2 Question 4

4. Let $\vec{α} = 3 \hat{i} + \hat{j}$ and $\vec{β} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ . If $\vec{β} = {\vec{β}}_{1} - {\vec{β}}_{2}$ , where ${\vec{β}}_{1}$ is parallel to $\vec{α}$ and ${\vec{β}}_{2}$ is perpendicular to $\vec{α}$ , then ${\vec{β}}_{1} \times {\vec{β}}_{2}$ is equal to

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Vectors 2 Question 4

4. Let α→=3i^+j^ and β→=2i^−j^+3k^. If β→=β→1−β→2, where β→1 is parallel to α→ and β→2 is perpendicular to α→, then β→1×β→2 is equal to

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4. Let $\vec{α} = 3 \hat{i} + \hat{j}$ and $\vec{β} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ . If $\vec{β} = {\vec{β}}_{1} - {\vec{β}}_{2}$ , where ${\vec{β}}_{1}$ is parallel to $\vec{α}$ and ${\vec{β}}_{2}$ is perpendicular to $\vec{α}$ , then ${\vec{β}}_{1} \times {\vec{β}}_{2}$ is equal to