SATHEE: Vectors 3 Question 29

Vectors 3 Question 29

29. Let $\vec{u}$ and $\vec{v}$ be unit vectors. If $\vec{w}$ is a vector such that $\vec{w} + (\vec{w} \times \vec{u}) = \vec{v}$ , then prove that $| (\vec{u} \times \vec{v}) \cdot \vec{w} | \leq \frac{1}{2}$ and that the equality holds if and only if $\vec{u}$ is perpendicular to $\vec{v}$ .

$(1999, 10 M)$

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Solution:

Given equation is $\vec{w} + (\vec{w} \times \vec{u}) = \vec{v}$

Taking cross product with $\vec{u}$ , we get

$\begin{array}{rr} \vec{u} \times [\vec{w} + (\vec{w} \times \vec{u})] = \vec{u} \times \vec{v} \\ \Rightarrow & \vec{u} \times \vec{w} + \vec{u} \times (\vec{w} \times \vec{u}) = \vec{u} \times \vec{v} \\ \Rightarrow & \vec{u} \times \vec{w} + (\vec{u} \cdot \vec{u}) \vec{w} - (\vec{u} \cdot \vec{w}) \vec{u} = \vec{u} \times \vec{v} \\ \Rightarrow & \vec{u} \times \vec{w} + \vec{w} - (\vec{u} \cdot \vec{w}) \vec{u} = \vec{u} \times \vec{v} \end{array}$

Now, taking dot product of Eq. (i) with $\vec{u}$ , we get

$\vec{u} \cdot \vec{w} + \vec{u} \cdot (\vec{w} \times \vec{u}) = \vec{u} \cdot \vec{v}$

$\Rightarrow \vec{u} \cdot \vec{w} = \vec{u} \cdot \vec{v} [∵ \vec{u} \cdot (\vec{w} \times \vec{u}) = 0]$

Now, taking dot product of Eq. (i) with $\vec{u}$ , we get $\vec{v} \cdot \vec{w} + \vec{v} \cdot (\vec{w} \times \vec{u}) = \vec{v} \cdot \vec{v}$

$\Rightarrow \vec{v} \cdot \vec{w} + [\vec{v} \vec{w} \vec{u}] = 1 \Rightarrow \vec{v} \cdot \vec{w} + [\vec{v} \vec{w} \vec{u}] - 1 = 0$

$\Rightarrow - (\vec{u} \times \vec{v}) \cdot \vec{w} - \vec{v} \cdot \vec{w} + 1 = 0$

$\Rightarrow 1 - \vec{v} \cdot \vec{w} = (\vec{u} \times \vec{v}) \cdot \vec{w}$

Taking dot product of Eq. (ii) with $\vec{w}$ , we get $(\vec{u} \times \vec{w}) \cdot \vec{w} + \vec{w} \cdot \vec{w} - (\vec{u} \cdot \vec{w}) (\vec{u} \cdot \vec{w}) = (\vec{u} \times \vec{v}) \cdot \vec{w}$

$\begin{array}{ll} \Rightarrow & 0 + | \vec{w} |^{2} - (\vec{u} \cdot \vec{w})^{2} = (\vec{u} \times \vec{v}) \cdot \vec{w} \\ \Rightarrow & (\vec{u} \times \vec{v}) \cdot \vec{w} = | \vec{w} |^{2} - (\vec{u} \cdot \vec{w})^{2} \end{array}$

Taking dot product of Eq. (i) with $\vec{w}$ , we get

$\begin{aligned} \vec{w} \cdot \vec{w} + (\vec{w} \times \vec{u}) \cdot \vec{w} = \vec{v} \cdot \vec{w} \\ \Rightarrow & | \vec{w} |^{2} + 0 = \vec{v} \cdot \vec{w} \\ \Rightarrow & | \vec{w} |^{2} = 1 - (\vec{u} \times \vec{v}) \cdot \vec{w} [from Eq. (iv)] \end{aligned}$

Again, from Eq. (v), we get

$(\vec{u} \times \vec{v}) \cdot \vec{w} + | \vec{w} |^{2} - (\vec{u} \cdot \vec{w})^{2} = 1 - (\vec{u} \times \vec{v}) \cdot \vec{w} - (\vec{u} \cdot \vec{w})^{2}$

$\Rightarrow 2 (\vec{u} \times \vec{v}) \cdot \vec{w} = 1 - (\vec{u} \cdot \vec{v})^{2}$

[from Eq. (iii)]

$\Rightarrow | (\vec{u} \times \vec{v}) \cdot \vec{w} | = \frac{1}{2} | 1 - (\vec{u} \cdot \vec{v})^{2} | \leq \frac{1}{2} [∵ (\vec{u} \cdot \vec{v})^{2} \geq 0]$

The equality holds if and only if $\vec{u} \cdot \vec{v} = 0$ or iff $\vec{u}$ is perpendicular to $\vec{v}$ .

Vectors 3 Question 29

29. Let $\vec{u}$ and $\vec{v}$ be unit vectors. If $\vec{w}$ is a vector such that $\vec{w} + (\vec{w} \times \vec{u}) = \vec{v}$ , then prove that $| (\vec{u} \times \vec{v}) \cdot \vec{w} | \leq \frac{1}{2}$ and that the equality holds if and only if $\vec{u}$ is perpendicular to $\vec{v}$ .

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Vectors 3 Question 29

29. Let u→ and v→ be unit vectors. If w→ is a vector such that w→+(w→×u→)=v→, then prove that |(u→×v→)⋅w→|≤12 and that the equality holds if and only if u→ is perpendicular to v→.

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29. Let $\vec{u}$ and $\vec{v}$ be unit vectors. If $\vec{w}$ is a vector such that $\vec{w} + (\vec{w} \times \vec{u}) = \vec{v}$ , then prove that $| (\vec{u} \times \vec{v}) \cdot \vec{w} | \leq \frac{1}{2}$ and that the equality holds if and only if $\vec{u}$ is perpendicular to $\vec{v}$ .