Vector Algebra

Given,

$\begin{array}{l}P(2\overrightarrow{a}+\overrightarrow{b})i.e,OP=2\overrightarrow{a}+\overrightarrow{b}\\ Q(\overrightarrow{a}-3\overrightarrow{b})i.e,OQ=\overrightarrow{a}-3\overrightarrow{b}\end{array}$

It is given that point R divides a line segment joining two points P and Q.

externally in the ratio 1:2 Then,

$\begin{array}{l}\overrightarrow{OR}=\frac{2(2\overrightarrow{a}+\overrightarrow{b})-(\overrightarrow{a}-3\overrightarrow{b})}{2-1}\\ =\frac{4\overrightarrow{a}+\overrightarrow{2b}-\overrightarrow{a}+3\overrightarrow{b}}{1}\\ \overrightarrow{OR}=3\overrightarrow{a}+5\overrightarrow{b}\end{array}$

$\therefore$ Position vector of the mid-point of RQ.

$\begin{array}{l}=\frac{\overrightarrow{OQ}+\overrightarrow{OR}}{2}\\ =\frac{(\overrightarrow{a}-3\overrightarrow{b})+(3\overrightarrow{a}+5\overrightarrow{b})}{2}\\ =\frac{\overrightarrow{a}-3\overrightarrow{b}+3\overrightarrow{a}+5\overrightarrow{b}}{2}\\ =\frac{4\overrightarrow{a}-2\overrightarrow{b}}{2}=2\overrightarrow{a}-\overrightarrow{b}=\overrightarrow{OR}Henceproved\end{array}$

Given,

$\begin{array}{l}A(1,-2,-8)\\ B(5,0,-2)\\ C(11,3,7)\end{array}$

Now,

Thus, A,B and C are collinear.

Let, $\lambda :1$ be the ratio that point B divides AC.

We have,

$\begin{array}{l}\overrightarrow{OB}=\frac{\lambda \overrightarrow{OC}+\overrightarrow{OA}}{\lambda +1}\\ 5\hat{i}-2\hat{k}=\frac{\lambda (11\hat{i}+3\hat{j}+7\hat{k})+(\hat{i}-2\hat{j}-8\widehat{k)}}{\lambda +1}\\ (5\hat{i}-2\hat{k})(\lambda +1)=11\lambda \hat{i}+3\lambda \hat{j}+7\lambda \hat{k}+\hat{i}-2\hat{j}-8\hat{k}\\ 5(\lambda +1)\hat{i}-2(\lambda +1)\hat{k}=(11\lambda +1)\hat{i}+(3\lambda -2)\hat{j}+(7\lambda -8)\hat{k}\end{array}$

On eQ.uating the corresponding component , we get

$\begin{array}{l}5(\lambda +1)=11\lambda +1\\ 5\lambda +5=11\lambda +1\\ 5-1=11\lambda -5\lambda \\ 4=6\lambda \\ \therefore \lambda =\frac{4}{6}=\frac{2}{3}\end{array}$

Hence, point B divides AC in the ratio $2:3$

Given,

$\begin{array}{l}\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}\\ \overrightarrow{b}=\widehat{2i}-\hat{j}+3\hat{k}\\ \overrightarrow{c}=\hat{i}-2\hat{j}+\hat{k}\end{array}$

Then,

We know,

$\begin{array}{l}\overrightarrow{a}=2\hat{i}+3\hat{j}-\hat{k}\\ \overrightarrow{b}=\hat{i}-2\hat{j}+\hat{k}\end{array}$

Let,  $\overrightarrow{c}$ be the resultant of $\overrightarrow{a}$ and $\overrightarrow{b}$

Then,

Given,

$x\left(\hat{i}+\hat{j}+\hat{k}\right)$ is a unit vector.

So,  $\left|x\left(\hat{i}+\hat{j}+\hat{k}\right)\right|=1$

Now,  $\left|x\left(\hat{i}+\hat{j}+\hat{k}\right)\right|=1$$\begin{array}{l}\\ \end{array}$

Let us take a $?ABC$ , which $\overrightarrow{CB}=\overrightarrow{a},\overrightarrow{CA}=\overrightarrow{b}\&\overrightarrow{AB}=\overrightarrow{c}$

So, by triangle law of vector addition, we have $\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}$

And, we know that $\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\&\left|\overrightarrow{c}\right|$ represent, the sides of $?ABC$

Also, it is known that the sum of the length of any slides of a triangle is greater than the third side. $\left|\overrightarrow{a}\right|<\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|$

Hence, it is not true that $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|$

$\therefore$ The girl's displacement from her initial point of departure is $=\frac{-5}{3}\hat{i}+\frac{\mathrm{3\surd 3}}{2}\hat{j}$

Given,

Point $P(x_1,y_1,z_1)\&Q(x_2,y_2,z_2)$

$\overrightarrow{PQ}$ = Position vector of Q.- Position vector of P

$=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$

Let $\overrightarrow{r}$ be unit vector in the XY-plane then,  $\overrightarrow{r}=cos\theta \hat{i}+sin\theta \hat{j}$

$\theta$ is the angle made by the unit vector with the positive direction of the X-axis.

Then,  $\theta =30^0$

$\therefore$ ReQ.uired unit vector $=\frac{}{2}\hat{i}+\frac{1}{2}\hat{j}$

(c) Given,

$\begin{array}{l}A=-\widehat{i}+\frac{1}{2}\widehat{j}+4\widehat{k}\\ B=\widehat{i}+\frac{1}{2}\widehat{j}+4\widehat{k}\\ C=\widehat{i}-\frac{1}{2}\widehat{j}+4\widehat{k}\\ D=-\widehat{i}-\frac{1}{2}\widehat{j}+4\widehat{k}\\ A\overrightarrow{B}=\left(1+1\right)\widehat{i}+\left(\frac{1}{2}-\frac{1}{2}\right)\widehat{j}+\left(4-4\right)\widehat{k}\\ =2\widehat{i}\\ B\overrightarrow{C}=\left(1-1\right)\widehat{i}+\left(-\frac{1}{2}-\frac{1}{2}\right)\widehat{j}+\left(4-4\right)\widehat{k}\\ =-\widehat{j}\end{array}$