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Vector Algebra Part 3

Section: Mathematics

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Question : 70 of 72

Marks: +1, -0

Let a unit vector

\hat{u}=x\hat{\dot{i}}+y\hat{j}+z\hat{k}

\;\frac{\pi}{2},\;\frac{\pi}{3}

\;\frac{2\pi}{3}

with the vectors

\;\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k},\;\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}

\;\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}\;\;

respectively. If

\vec{v}=\;\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}

\left|\hat{u}-\vec{v}\right|^2

[29-Jan-2024 Shift 2]

$\;\frac{11}{2}$
$\;\frac{5}{2}$
9
7

Solution: 👈: Video Solution

Unit vector

\hat{u}=x\hat{i}+y\hat{j}+z\hat{k}

\;\vec{p}_1=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}, \vec{p}_2=\frac{1}{\sqrt{2}}\hat{j}

+\frac{1}{\sqrt{2}}\hat{k}

\;\vec{p}_3=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}

Now angle between

\hat{u}

and

\vec{p}_1=\;\frac{\pi}{2}

\hat{u}\cdot\vec{p}_1=0 \Rightarrow \;\frac{x}{\sqrt{2}}+\frac{z}{\sqrt{2}}=0

\Rightarrow x+z=0

..............(i)

Angle between

\hat{u}

and

\vec{p}_2=\;\frac{\pi}{3}

\hat{u}\cdot\vec{p}_2 = \left|\hat{u}\right| \cdot \left|\vec{p}_2\right| \cos \;\frac{\pi}{3}

\Rightarrow \;\frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2}

\Rightarrow y+z=\;\frac{1}{\sqrt{2}}

.............(ii)

Angle between

\hat{u}

and

\vec{p}_3=\;\frac{2\pi}{3}

\hat{u}\cdot\vec{p}_3 = \left|\hat{u}\right| \cdot \left|\vec{p}_3\right| \cos \;\frac{2\pi}{3}

\Rightarrow \;\frac{x}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{-1}{2}

\Rightarrow x+y=\;\frac{-1}{\sqrt{2}}

.............(iii)

from equation (i), (ii) and (iii) we get

x=\;\frac{-1}{\sqrt{2}}\;\;y=0\;\;z=\;\frac{1}{\sqrt{2}}

\;\;\text{Thus}\;\;\hat{u}-\vec{v}=\frac{-1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}-\frac{1}{\sqrt{2}}\hat{i}

-\frac{1}{\sqrt{2}}\hat{j}-\frac{1}{\sqrt{2}}\hat{k}

\hat{u}-\vec{v}=\;\frac{-2}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}

\therefore \left|\hat{u}-\vec{v}\right|^2 = \left(\sqrt{\;\frac{4}{2}+\;\frac{1}{2}}\right)^2

=\;\frac{5}{2}

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