SATHEE: Vectors 4 Question 12

Vectors 4 Question 12

12. Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors having magnitudes 1 , 1 and 2 , respectively. If $\vec{a} \times (\vec{a} \times \vec{c}) + \vec{b} = \vec{0}$ , then the actue angle between $\vec{a}$ and $\vec{c}$ is

(1997, 2M)

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Answer:

Correct Answer: 12. $\frac{π}{6}$

Solution:

Given,

$\vec{a} \times (\vec{a} \times \vec{c}) + \vec{b} = \vec{0}$

$\begin{array}{ll} \Rightarrow & (\vec{a} \cdot \vec{c}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{c} + \vec{b} = \vec{0} \\ \Rightarrow & (2 \cos θ) \vec{a} - \vec{c} + \vec{b} = \vec{0} \end{array}$

$\begin{array}{lrl} \Rightarrow & (2 \cos θ \vec{a} - \vec{c})^{2} = (- \vec{b})^{2} \\ \Rightarrow & 4 \cos^{2} θ \cdot | \vec{a} |^{2} + | \vec{c} |^{2} - 2 \cdot 2 \cos θ \vec{a} \cdot \vec{c} = | \vec{b} |^{2} \\ \Rightarrow & 4 \cos^{2} θ - 4 - 8 \cos^{2} θ = 1 \\ \Rightarrow & 4 \cos^{2} θ = 3 \\ \Rightarrow & \cos θ = \pm \frac{\sqrt{3}}{2} \end{array}$

For $θ$ to be acute,

$\cos θ = \frac{\sqrt{3}}{2} \Rightarrow θ = \frac{π}{6}$

Vectors 4 Question 12

12. Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors having magnitudes 1 , 1 and 2 , respectively. If $\vec{a} \times (\vec{a} \times \vec{c}) + \vec{b} = \vec{0}$ , then the actue angle between $\vec{a}$ and $\vec{c}$ is

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Vectors 4 Question 12

12. Let a→,b→ and c→ be three vectors having magnitudes 1 , 1 and 2 , respectively. If a→×(a→×c→)+b→=0→, then the actue angle between a→ and c→ is

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12. Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors having magnitudes 1 , 1 and 2 , respectively. If $\vec{a} \times (\vec{a} \times \vec{c}) + \vec{b} = \vec{0}$ , then the actue angle between $\vec{a}$ and $\vec{c}$ is