SATHEE: Vectors 2 Question 22

Vectors 2 Question 22

22. For any two vectors $\vec{u}$ and $\vec{v}$ , prove that

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

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

$(1998, 8 M)$

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Solution:

(i) Since, $\vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos θ$ and $\vec{u} \times \vec{v} = | \vec{u} | | \vec{v} | \sin θ \hat{n}$ where, $θ$ is the angle between $\vec{u}$ and $\vec{v}$ and $\hat{n}$ is unit vector perpendicular to the plane of $\vec{u}$ and $\vec{v}$ .

Again,

$| \vec{u} \cdot \vec{v} |^{2} = | \vec{u} |^{2} | \vec{v} |^{2} \cos^{2} θ and$

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

$∴ | \vec{u} \cdot \vec{v} |^{2} + | \vec{u} \times \vec{v} |^{2} = | \vec{u} |^{2} | \vec{v} |^{2} (\cos^{2} θ + \sin^{2} θ)$

$= | \vec{u} |^{2} | \vec{v} |^{2}$

(ii) $| \vec{u} + \vec{v} + (\vec{u} \times \vec{v}) |^{2}$

$\begin{aligned} = | \vec{u} + \vec{v} |^{2} + | \vec{u} \times \vec{v} |^{2} + 2 (\vec{u} + \vec{v}) \cdot (\vec{u} \times \vec{v}) \\ = | \vec{u} |^{2} + | \vec{v} |^{2} + 2 \vec{u} \cdot \vec{v} + | \vec{u} \times \vec{v} |^{2} + 0 \end{aligned}$

$[∵ \vec{u} \times \vec{v}$ is perpendicular to the plane of $\vec{u}$ and $\vec{v}]$

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

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