Question

plz ans this asap class 11 physics

Answer 1

Answer:

if right then may help u mate....

Answer 2

Given:

$\rm{\vec{A}=3\hat{i}-4\hat{j}+5\hat{k}}$

$\rm{\vec{B}=2\hat{i}+3\hat{j}-4\hat{k}}$

To Find:

A unit vector parallel to $\rm{\vec{A}+\vec{B}}$

Solution:

We know that,

• A unit vector parallel to given vector $\rm{\vec{R}}$ is given by

$\pink{\boxed{\rm{\hat{R}=\dfrac{\vec{R}}{\mid\vec{R}\mid}}}}$

where

$\rm{\mid\vec{R}\mid}$ is the magnitude of vector $\rm{\vec{R}}$

• Magnitude of given vector $\rm{\vec{R}=x\hat{i}+y\hat{j}+z\hat{k}}$ is given by

$\purple{\boxed{\rm{\mid\vec{R}\mid=\sqrt{(x)^{2}+(y)^{2}+(z)^{2}}}}}$

$\rule{190}{1}$

Let $\rm{\vec{P}=\vec{A}+\vec{B}}$

So,

$\longrightarrow\rm{\vec{P}=3\hat{i}-4\hat{j}+5\hat{k}+2\hat{i}+3\hat{j}-4\hat{k}}$

$\longrightarrow\rm{\vec{P}=(3+2)\hat{i}+(-4+3)\hat{j}+(5-4)\hat{k}}$

$\longrightarrow\rm{\vec{P}=(5)\hat{i}+(-1)\hat{j}+(1)\hat{k}}$

$\longrightarrow\rm{\vec{P}=5\hat{i}-\hat{j}+\hat{k}}$

$\rule{190}{1}$

Now, we have to find unit vector parallel to $\rm{\vec{P}}$

Let the unit vector in the direction of $\rm{\vec{P}}$ be $\rm{\hat{P}}$

So,

$\longrightarrow\rm{\hat{P}=\dfrac{\vec{P}}{\mid\vec{P}\mid}}$

$\longrightarrow\rm{\hat{P}=\dfrac{5\hat{i}-\hat{j}+\hat{k}}{\sqrt{(5)^{2}+(-1)^{2}+(1)^{2}}}}$

$\longrightarrow\rm{\hat{P}=\dfrac{5\hat{i}-\hat{j}+\hat{k}}{\sqrt{25+1+1}}}$

$\longrightarrow\rm{\hat{P}=\dfrac{5\hat{i}-\hat{j}+\hat{k}}{\sqrt{27}}}$

$\longrightarrow\rm\green{\hat{P}=\dfrac{5}{\sqrt{27}}\hat{i}-\dfrac{1}{\sqrt{27}}\hat{j}+\dfrac{1}{\sqrt{27}}\hat{k}}$

Hence, $\rm{\hat{P}}$ is the required unit vector parallel to $\rm{\vec{A}+\vec{B}}$.