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QJEEAdv21Ma06bc: JEE Advanced 2021 Physics - Class 11 JEEAdv21Ma06bc Solution: Let , and be vectors in three-dimensional space, where and are…

Let u→\overrightarrow u , v→\overrightarrow v and w→\overrightarrow w be vectors in three-dimensional space, where u→\overrightarrow u and v→\overrightarrow v are unit vectors which are not perpendicular to each other and u→\overrightarrow u . w→\overrightarrow w = 1, v→\overrightarrow v . w→\overrightarrow w = 1, w→\overrightarrow w . w→\overrightarrow w = 4

If the volume of the paralleopiped, whose adjacent sides are represented by the vectors, u→\overrightarrow u , v→\overrightarrow v and w→\overrightarrow w , is 2\sqrt 2 , then the value of ∣3u→+5v→∣\left| {3\overrightarrow u + 5\overrightarrow v } \right| is ___________.

Step-by-Step Verified Solution

Given: ∣u⃗∣=1,∣v⃗∣=1|\vec{u}|=1,|\vec{v}|=1

u⃗\vec{u} is not perpendicular to v⃗\vec{v} u⃗⋅v⃗≠0\vec{u} \cdot \vec{v} \neq 0

And u⃗⋅w⃗=1v⃗⋅w⃗=1w⃗⋅w⃗=4\vec{u} \cdot \vec{w}=1 \quad \vec{v} \cdot \vec{w}=1 \quad \vec{w} \cdot \vec{w}=4

amp;⇒w⃗⋅w⃗=∣w⃗∣2=4amp;⇒∣w⃗∣=2\begin{aligned}& \Rightarrow \vec{w} \cdot \vec{w}=|\vec{w}|^2=4 \\\\& \Rightarrow|\vec{w}|=2\end{aligned}

Volume of parallelepiped with u⃗,v⃗\vec{u}, \vec{v} and w⃗\vec{w} as its sides,

amp;=[u⃗v⃗w⃗]=2amp; Now, [u⃗v⃗w⃗]2=∣u⃗⋅u⃗amp;u⃗⋅v⃗amp;u⃗⋅w⃗v⃗⋅u⃗amp;v⃗⋅v⃗amp;v⃗⋅w⃗w⃗⋅u⃗amp;w⃗⋅v⃗amp;w⃗⋅w⃗∣=2amp;⇒∣1amp;u⃗⋅v⃗amp;1v⃗⋅u⃗amp;1amp;11amp;1amp;4∣=2\begin{aligned}& =[\vec{u} \vec{v} \vec{w}]=\sqrt{2} \\\\& \text { Now, }[\vec{u} \vec{v} \vec{w}]^2=\left|\begin{array}{ccc}\vec{u} \cdot \vec{u} & \vec{u} \cdot \vec{v} & \vec{u} \cdot \vec{w} \\\vec{v} \cdot \vec{u} & \vec{v} \cdot \vec{v} & \vec{v} \cdot \vec{w} \\\vec{w} \cdot \vec{u} & \vec{w} \cdot \vec{v} & \vec{w} \cdot \vec{w}\end{array}\right|=2 \\\\& \Rightarrow\left|\begin{array}{ccc}1 & \vec{u} \cdot \vec{v} & 1 \\\vec{v} \cdot \vec{u} & 1 & 1 \\1 & 1 & 4\end{array}\right|=2\end{aligned}

amp;⇒1(4−1)−u⃗⋅v⃗(4u⃗⋅v⃗−1)+1(u⃗⋅v⃗−1)=2amp;⇒3−4(u⃗v⃗)2+u⃗v⃗+u⃗v⃗−1=2amp;⇒−4(u⃗v⃗)2+2u⃗v⃗+2=2amp;⇒−4(u⃗v⃗)2+2u⃗v⃗=0amp;⇒2u⃗v⃗(−2u⃗v⃗+1)=0amp;⇒u⃗⋅v⃗=12,u⃗⋅v⃗≠0amp; Now, ∣3u⃗+5v⃗∣=9+25+30(12)amp;⇒∣3u⃗+5v⃗∣=49amp;⇒∣3u⃗+5v⃗∣=7\begin{aligned}& \Rightarrow 1(4-1)-\vec{u} \cdot \vec{v}(4 \vec{u} \cdot \vec{v}-1)+1(\vec{u} \cdot \vec{v}-1)=2 \\\\& \Rightarrow 3-4(\vec{u} \vec{v})^2+\vec{u} \vec{v}+\vec{u} \vec{v}-1=2 \\\\& \Rightarrow-4(\vec{u} \vec{v})^2+2 \vec{u} \vec{v}+2=2 \\\\& \Rightarrow-4(\vec{u} \vec{v})^2+2 \vec{u} \vec{v}=0 \\\\& \Rightarrow 2 \vec{u} \vec{v}(-2 \vec{u} \vec{v}+1)=0 \\\\& \Rightarrow \vec{u} \cdot \vec{v}=\frac{1}{2}, \vec{u} \cdot \vec{v} \neq 0 \\\\& \text { Now, }|3 \vec{u}+5 \vec{v}|=\sqrt{9+25+30\left(\frac{1}{2}\right)} \\\\& \Rightarrow|3 \vec{u}+5 \vec{v}|=\sqrt{49} \\\\& \Rightarrow|3 \vec{u}+5 \vec{v}|=7\end{aligned}

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