Let a unit vector hat{u}=x hat{i}+y hat{j}+z | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given unit vector $\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$.Let $\overrightarrow{p}_1=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}$,$\overr

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(B) Given unit vector $\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$.Let $\overrightarrow{p}_1=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}$,$\overrightarrow{p}_2=\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$,and $\overrightarrow{p}_3=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$.Since $\hat{u} \cdot \overrightarrow{p}_1 = |\hat{u}| |\overrightarrow{p}_1| \cos(\frac{\pi}{2}) = 0$,we have $\frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0 \Rightarrow x+z=0$ $(i)$.Since $\hat{u} \cdot \overrightarrow{p}_2 = |\hat{u}| |\overrightarrow{p}_2| \cos(\frac{\pi}{3}) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}$,we have $\frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2} \Rightarrow y+z = \frac{1}{\sqrt{2}}$ $(ii)$.Since $\hat{u} \cdot \overrightarrow{p}_3 = |\hat{u}| |\overrightarrow{p}_3| \cos(\frac{2\pi}{3}) = 1 \cdot 1 \cdot (-\frac{1}{2}) = -\frac{1}{2}$,we have $\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = -\frac{1}{2} \Rightarrow x+y = -\frac{1}{\sqrt{2}}$ $(iii)$.Adding $(i), (ii), (iii)$,we get $2(x+y+z) = 0 \Rightarrow x+y+z = 0$.Subtracting $(ii)$ from this,$x = -\frac{1}{\sqrt{2}}$. Subtracting $(iii)$ from this,$z = \frac{1}{\sqrt{2}}$. Subtracting $(i)$ from this,$y = 0$.Thus $\hat{u} = -\frac{1}{\sqrt{2}} \hat{i} + 0 \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$.Then $\hat{u}-\overrightarrow{v} = (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) \hat{i} + (0 - \frac{1}{\sqrt{2}}) \hat{j} + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) \hat{k} = -\frac{2}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} = -\sqrt{2} \hat{i} - \frac{1}{\sqrt{2}} \hat{j}$.$|\hat{u}-\overrightarrow{v}|^2 = (-\sqrt{2})^2 + (-\frac{1}{\sqrt{2}})^2 = 2 + \frac{1}{2} = \frac{5}{2}$.

Similar Questions

If $|a| = 2$,$|b| = 5$ and $|a \times b| = 8$,then $a \cdot b$ is equal to

If $\vec{a}, \vec{b}, \vec{c}$ are vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=7, |\vec{b}|=5, |\vec{c}|=3$,then the angle between vector $\vec{b}$ and $\vec{c}$ is: (in $^{\circ}$)

If the position vectors of three points $A, B, C$ are $\hat{i}+2\hat{j}+\hat{k}$,$2\hat{i}-\hat{j}+2\hat{k}$ and $\hat{i}+\hat{j}+2\hat{k}$ respectively,then the perpendicular distance of the point $C$ from the line $AB$ is

Let $\vec{a}, \vec{b}, \vec{c}$ be three unit vectors such that $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} - \vec{a} \cdot \vec{c} = \frac{3}{2}$. Then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module