Let vec{a}=x hat{i}+y hat{j}+z hat{k} and x=2 | vedclass

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

Question

(C) Given $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$ and $x=2 y$. $|\vec{a}| = \sqrt{x^2+y^2+z^2} = \sqrt{(2y)^2+y^2+z^2} = \sqrt{5y^2+z^2}$. Given $|\ve

Vedclass · Accepted Answer

$2 \sqrt{5} \hat{i}+\sqrt{5} \hat{j}-5 \hat{k}$

Vedclass · Answer

(C) Given $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$ and $x=2 y$. $|\vec{a}| = \sqrt{x^2+y^2+z^2} = \sqrt{(2y)^2+y^2+z^2} = \sqrt{5y^2+z^2}$. Given $|\vec{a}| = 5 \sqrt{2}$,so $5y^2+z^2 = (5 \sqrt{2})^2 = 50$. Since $\vec{a}$ makes an angle of $135^{\circ}$ with the $z$-axis,the component along the $z$-axis is $z = |\vec{a}| \cos 135^{\circ}$. $z = 5 \sqrt{2} 	imes (-\frac{1}{\sqrt{2}}) = -5$. Substituting $z = -5$ into the equation $5y^2+z^2 = 50$: $5y^2 + (-5)^2 = 50 \Rightarrow 5y^2 + 25 = 50 \Rightarrow 5y^2 = 25 \Rightarrow y^2 = 5 \Rightarrow y = \pm \sqrt{5}$. Since $x = 2y$,we have $x = \pm 2 \sqrt{5}$. Thus,$\vec{a} = \pm 2 \sqrt{5} \hat{i} \pm \sqrt{5} \hat{j} - 5 \hat{k}$. Comparing with the options,the correct vector is $2 \sqrt{5} \hat{i} + \sqrt{5} \hat{j} - 5 \hat{k}$.

Let $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$ and $x=2 y$. If $|\vec{a}|=5 \sqrt{2}$ and $\vec{a}$ makes an angle of $135^{\circ}$ with the $z$-axis,then $\vec{a}=$

Similar Questions

The position vectors of the points $A, B, C$ are $(2\hat{i}+\hat{j}-\hat{k}), (3\hat{i}-2\hat{j}+\hat{k})$ and $(\hat{i}+4\hat{j}-3\hat{k})$ respectively. These points

Let $\vec{a}$ and $\vec{b}$ be two non-collinear vectors. For what values of $x$ and $y$ is the equation $2\vec{u} - \vec{v} = \vec{w}$ true,where $\vec{u} = x\vec{a} + 2y\vec{b}$,$\vec{v} = -2y\vec{a} + 3x\vec{b}$,and $\vec{w} = 4\vec{a} - 2\vec{b}$?

If $\vec{a}=\vec{b}+\vec{c},$ then is it true that $|\vec{a}|=|\vec{b}|+|\vec{c}|$ ? Justify your answer.

If $\vec{a}$ is any vector in space,then

The perimeter of a triangle with sides represented by vectors $\vec{a} = 3\hat{i} + 4\hat{j} + 5\hat{k}$,$\vec{b} = 4\hat{i} - 3\hat{j} - 5\hat{k}$,and $\vec{c} = 7\hat{i} + \hat{j}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module