SATHEE: Vectors 2 Question 13

Vectors 2 Question 13

13. Let $\vec{a}$ and $\vec{b}$ be two non-collinear unit vectors. If $\vec{u} = \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}$ and $\vec{v} = \vec{a} \times \vec{b}$ , then $| \vec{v} |$ is

(1999, 3M)

(a) $| \vec{u} |$

(b) $| \vec{u} | + | \vec{u} \cdot \vec{a} |$

(d) $| \vec{u} | + \vec{u} \cdot (\vec{a} + \vec{b})$

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Answer:

Correct Answer: 13. (c)

Solution:

Let $θ$ be the angle between $\vec{a}$ and $\vec{b}$ . Since, $\vec{a}$ and $\vec{a}$ are non-collinear vectors, then $θ \neq 0$ and $θ \neq π$ .

We have, $\vec{a} \cdot \vec{b} = | \vec{a} | | \vec{a} | \cos θ$

$= \cos θ [∵ | \vec{a} | = 1, | \vec{b} | = 1, given]$

Now, $\vec{u} = \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} \Rightarrow | \vec{u} | = | \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} |$

$\Rightarrow | \vec{u} |^{2} = | \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} |^{2}$

$\Rightarrow | \vec{u} |^{2} = | \vec{a} - \cos θ \vec{b} |^{2}$

$\Rightarrow | \vec{u} |^{2} = | \vec{a} |^{2} + \cos^{2} θ | \vec{b} |^{2} - 2 \cos θ (\vec{a} \cdot \vec{b})$

$\Rightarrow | \vec{u} |^{2} = 1 + \cos^{2} θ - 2 \cos^{2} θ$

$\Rightarrow | \vec{u} |^{2} = 1 - \cos^{2} θ \Rightarrow | \vec{u} |^{2} = \sin^{2} θ$

Also, $\vec{v} = \vec{a} \times \vec{b}$

[given]

$\Rightarrow | \vec{v} |^{2} = | \vec{a} \times \vec{b} |^{2} \Rightarrow | \vec{v} |^{2} = | \vec{a} |^{2} | \vec{b} |^{2} \cdot \sin^{2} θ$

$\Rightarrow | \vec{v} |^{2} = \sin^{2} θ ∴ | \vec{u} |^{2} = | \vec{v} |^{2}$

Now, $\vec{u} \cdot \vec{a} = [\vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}] \cdot \vec{a} = \vec{a} \cdot \vec{a} - (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{a})$

$= (\vec{a})^{2} - \cos^{2} θ = 1 - \cos^{2} θ = \sin^{2} θ$

$\begin{aligned} ∴ | \vec{u} | + | \vec{u} \cdot \vec{a} | & = \sin θ + \sin^{2} θ \neq | \vec{v} | \\ \vec{u} \cdot \vec{b} & = [\vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}] \cdot \vec{b} \\ = \vec{a} \cdot \vec{b} - (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{b}) = \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{b} | \vec{b} |^{2} \\ = \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{b} = 0 \end{aligned}$

$∴ | \vec{u} | + | \vec{u} \cdot \vec{b} | = | \vec{u} | + 0 = | \vec{u} | = | \vec{v} |$

Also, $\vec{u} \cdot (\vec{a} + \vec{b}) = \vec{u} \cdot \vec{a} + \vec{u} \cdot \vec{b} = \vec{u} \cdot \vec{a}$

$\Rightarrow | \vec{u} | + \vec{u} \cdot (\vec{a} + \vec{b}) = | \vec{u} | + \vec{u} \cdot \vec{a} \neq | \vec{v} |$

Vectors 2 Question 13

13. Let $\vec{a}$ and $\vec{b}$ be two non-collinear unit vectors. If $\vec{u} = \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}$ and $\vec{v} = \vec{a} \times \vec{b}$ , then $| \vec{v} |$ is

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Vectors 2 Question 13

13. Let a→ and b→ be two non-collinear unit vectors. If u→=a→−(a→⋅b→)b→ and v→=a→×b→, then |v→| is

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13. Let $\vec{a}$ and $\vec{b}$ be two non-collinear unit vectors. If $\vec{u} = \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}$ and $\vec{v} = \vec{a} \times \vec{b}$ , then $| \vec{v} |$ is