Vector Algebra

$(2\hat{i}+4\hat{j}-5\hat{k})+(\lambda \hat{i}+2\hat{j}+3\hat{k})=(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}$

The unit vector along $(2\hat{i}+4\hat{j}-5\hat{k})+(\lambda \hat{i}+2\hat{j}+3\hat{k})$ is given as;

By Q.uestion, scalar product of $\left(\hat{i}+\hat{j}+\hat{k}\right)$ with this unit vector is 1.

Given,

$\begin{array}{l}\overrightarrow{a}=\hat{i}+4\hat{j}+2\hat{k}\\ \overrightarrow{b}=3\hat{i}-2\hat{j}+7\hat{k}\\ \overrightarrow{c}=2\hat{i}-\hat{j}+4\hat{k}\end{array}$

Let, $\overrightarrow{d}=d_1\hat{i}+d_2\hat{j}+d_3\hat{k}$

Since, $\overrightarrow{d}$ is perpendicular to both $\overrightarrow{a}\&\overrightarrow{b}$

$\begin{array}{l}\overrightarrow{d}.\overrightarrow{a}=0\\ \Rightarrow d_1+d_{2}4+d_{3}2=0-----(1)\\ \overrightarrow{d}.\overrightarrow{b}=0\\ \Rightarrow d_{1}3+d_2(-2)+d_3(7)=0\\ \Rightarrow d_{1}3-2d_2+7d_3=0-----(2)\end{array}$

We know,

$\begin{array}{l}\overrightarrow{c}.\overrightarrow{d}=15\\ \Rightarrow 2d_1-d_2+4d_3=15-----(3)\\ From,(1)\\ d_1+4d_2+2d_3=0\\ d_1=-4d_2-2d_3\end{array}$

Putting this value in (3) we get

$\begin{array}{l}\Rightarrow 2(-4d_2-2d_3)-d_2+4d_3=15\\ \Rightarrow -8d_2-4d_3-d_2+4d_3=15\\ \Rightarrow -9d_2=15\\ \Rightarrow d_2=\frac{15}{9}=-\frac{5}{3}\end{array}$

Putting $d_1\&d_2$ value in (2), we get

$\begin{array}{l}\Rightarrow 3d_1-2d_2+7d_3=0\\ \Rightarrow 3(-4d_2-2d_3)-2\left(-\frac{5}{3}\right)+7d_3=0\\ \Rightarrow -12*\left(-\frac{5}{3}\right)-6d_3+\frac{10}{3}+7d_3=0\\ \Rightarrow 20+d_3+\frac{10}{3}=0\\ \Rightarrow d_3=-20-\frac{10}{3}=\frac{-60-10}{3}=\frac{-70}{3}\\ Now,d_1=-4*-\frac{5}{3}-2*-\frac{70}{3}\\ =\frac{20}{3}+\frac{140}{3}=\frac{160}{3}\\ \therefore d_1=\frac{160}{3},d_2=-\frac{5}{3},d_3=\frac{-70}{3}\\ \overrightarrow{d}=\frac{160}{3}\hat{i}-\frac{5}{3}\hat{j}\frac{-70}{3}\hat{k}=\frac{1}{3}(160\hat{i}-5\hat{j}-70\hat{k})\end{array}$

$\therefore$ The reQ.uired vector is $\frac{1}{3}(160\hat{i}-5\hat{j}-70\hat{k})$

Kindly go through the solution

Given,

Adjacent sides of parallelogram are

$\begin{array}{l}\overrightarrow{a}=2\hat{i}-4\hat{j}+5\hat{k}\\ \overrightarrow{b}=\hat{i}-2\hat{j}-3\hat{k}\end{array}$

$\therefore$ Diagonal of parallelogram = $\overrightarrow{a}+\overrightarrow{b}$

$\begin{array}{l}\Rightarrow \overrightarrow{a}+\overrightarrow{b}=(2+1)\hat{i}+(-4+(-2))\hat{j}+(5+(-3))\hat{k}\\ =3\hat{i}-6\hat{j}+2\hat{k}\end{array}$

Thus, the unit vector parallel to diagonal

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 $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|$