SATHEE: Vectors 4 Question 13

Vectors 4 Question 13

13. If $\vec{A}, \vec{B}$ and $\vec{C}$ are vectors such that $| \vec{B} | = | \vec{C} |$ . Prove that $[(\vec{A} + \vec{B}) \times (\vec{A} + \vec{C})] \times (\vec{B} \times \vec{C}) \cdot (\vec{B} + \vec{C}) = \vec{0}$ .

(1997, 5M)

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Answer:

Correct Answer: 13. (i) $\vec{X} = (\frac{\vec{c}}{| \vec{A} |^{2}}) \vec{A} - (\frac{1}{| \vec{A} |^{2}}) (\vec{A} \times \vec{B})$ (ii) $(A_{2} \hat{i} - A_{1} \hat{j} + A_{3} \hat{k})$

Solution:

Now, $(\vec{A} + \vec{B}) \times (\vec{A} + \vec{C})$

$\begin{aligned} = & \vec{A} \times \vec{A} + \vec{B} \times \vec{A} + \vec{A} \times \vec{C} + \vec{B} \times \vec{C} \\ = & \vec{B} \times \vec{A} + \vec{A} \times \vec{C} + \vec{B} \times \vec{C} [∵ \vec{A} \times \vec{A} = 0] \\ ∴ [(\vec{A} + \vec{B}) \times & (\vec{A} + \vec{C})] \times (\vec{B} \times \vec{C}) \\ = & [\vec{B} \times \vec{A} + \vec{A} \times \vec{C} + \vec{B} \times \vec{C}] \times (\vec{B} \times \vec{C}) \\ = & (\vec{B} \times \vec{A}) \times (\vec{B} \times \vec{C}) + (\vec{A} \times \vec{C}) \times (\vec{B} \times \vec{C}) \\ = & (\vec{B} \times \vec{A}) \cdot \vec{C} \vec{B} - (\vec{B} \times \vec{A}) \cdot \vec{B} \vec{C} \\ + (\vec{A} \times \vec{C}) \cdot \vec{C} \vec{B} - (\vec{A} \times \vec{C}) \cdot \vec{B} \vec{C} \\ [∵ (\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a}] \\ = & [\vec{B} \vec{A} \vec{C}] \vec{B} - [\vec{A} \vec{C} \vec{B}] \vec{C} \end{aligned}$

$[∵ [\vec{a} \vec{b} \vec{c}] = 0$ , if any two of $\vec{a}, \vec{b}, \vec{c}$ are equal ] = [ $\vec{A} \vec{C} \vec{B}$ ] $(\vec{B} - \vec{C})$

Now, $[(\vec{A} + \vec{B}) \times (\vec{A} + \vec{C})] \times (\vec{B} \times \vec{C}) \cdot (\vec{B} + \vec{C})$

$\begin{aligned} = ([\vec{A} \vec{C} \vec{B}] \vec{B} - \vec{C}) \cdot (\vec{B} + \vec{C}) \\ = [\vec{A} \vec{C} \vec{B}] (\vec{B} - \vec{C}) \cdot (\vec{B} + \vec{C}) \\ = [\vec{A} \vec{C} \vec{B}] | \vec{B} |^{2} - | \vec{C} |^{2} = \vec{0} [∵ | \vec{B} | = | \vec{C} |, given] \end{aligned}$

Vectors 4 Question 13

13. If $\vec{A}, \vec{B}$ and $\vec{C}$ are vectors such that $| \vec{B} | = | \vec{C} |$ . Prove that $[(\vec{A} + \vec{B}) \times (\vec{A} + \vec{C})] \times (\vec{B} \times \vec{C}) \cdot (\vec{B} + \vec{C}) = \vec{0}$ .

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Vectors 4 Question 13

13. If A→,B→ and C→ are vectors such that |B→|=|C→|. Prove that [(A→+B→)×(A→+C→)]×(B→×C→)⋅(B→+C→)=0→.

Answer:

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13. If $\vec{A}, \vec{B}$ and $\vec{C}$ are vectors such that $| \vec{B} | = | \vec{C} |$ . Prove that $[(\vec{A} + \vec{B}) \times (\vec{A} + \vec{C})] \times (\vec{B} \times \vec{C}) \cdot (\vec{B} + \vec{C}) = \vec{0}$ .