Maths Solve the Following Question.(2 Marks)

Let $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ be the position vectors of points $A , B , C , D$ respectively
$\overline{A D}+\overline{B D}+\overline{C D}=\bar{O}$
$(\bar{d}-\bar{a})+(\bar{d}-\bar{b})+(\bar{d}-\bar{c})=\bar{O}$
$3 \bar{d}-(\bar{a}+\bar{b}+\bar{c})=\bar{O}$
$3 \bar{d}=\bar{a}+\bar{b}+\bar{c}$
$\bar{d}=\frac{\bar{a}+\bar{b}+\bar{c}}{3}$
$\bar{d}$ represents centroid of the triangle.
Point D is the centroid of the ΔABC.

The direction ratios of the line are 1 + 4, 3 − 2, −2 − 3 i.e., 5, 1, −5
∴ the direction cosines of the line are
$\begin{aligned} & \pm \frac{5}{\sqrt{5^2+1^2+(-5)^2}}, \pm \frac{1}{\sqrt{5^2+1^2+(-5)^2}}, \pm-\frac{5}{\sqrt{5^2+1^2+(-5)^2}} \\ & \text { i.e } \pm \frac{5}{\sqrt{51}}, \pm \frac{1}{\sqrt{51}}, \pm-\frac{5}{\sqrt{51}}\end{aligned}$

Let $\bar{a}$ and $\bar{b}$ be the position vectors of the points $A$ and $B$ respectively.
Then, $\bar{a}=2 \hat{i}-6 \hat{j}+8 \widehat{k}$ and $\bar{b}=-\hat{i}+3 \hat{j}-4 \widehat{k}$
Let R ($\bar{r}$) be the point which divides the line segment joining the points A and B externally in the ratio 1 : 3.

3
$\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{k} \times \hat{i})+\hat{k} \cdot(\hat{i} \times \hat{j})$
$=\hat{\imath} \cdot \hat{\imath}+\hat{j} \cdot \hat{\jmath}+\hat{k} \cdot \hat{k}$
= 1 + 1 + 1
= 3

Hence option (d)

Let $\bar{a}, \bar{b}, \& \bar{c}$ be position vectors of $A, B \& C$
$\overline{ AB }=\overline{ b }-\overline{ a }$
$=(-2 \hat{i}+\hat{j})-(-7 \hat{i}+4 \hat{j}-2 \hat{k})$
$=5 \hat{ i }-3 \hat{ j }+2 \hat{ k }$
$\overline{ AC }=\overline{ c }-\overline{ a }$
$=(3 \hat{ i }-2 \hat{ j }+2 \hat{ k })-(-7 \hat{ i }+4 \hat{ j }-2 \hat{ k })$
$=10 \hat{ i }-6 \hat{ j }+4 \hat{ k }$
$=2[5 \hat{ i }-3 \hat{ j }+2 \hat{ k }]$
$\Rightarrow \overline{ AC }=2 \overline{( AB )}$
$\Rightarrow \overline{ AC }$ is a scalar multiple of $\overline{ AB }$
$\Rightarrow \overline{ AC } \& \overline{ AB }$
∵ A is common
$\Rightarrow A B \& C$ are collinear.

If $\vec{a} \vec{b}$ and $\vec{c}$ are conterminus edges of parallelopiped then the volume of the parallelopiped $=[\vec{a} \vec{b} \vec{c}]$
where $\vec{a}=2 \hat{i}+5 \hat{j}-4 \widehat{k}$
$\vec{b}=5 \hat{i}+7 \hat{j}+5 \widehat{k}$
$\vec{c}=4 \hat{i}+5 \hat{j}-2 \widehat{k}$
$\therefore V=[\vec{a} \vec{b} \vec{c}]=\left[\begin{array}{ccc}2 & 5 & -4 \\ 5 & 7 & 5 \\ 4 & 5 & -2\end{array}\right]$
$=2(-14-25)-5(-10-20)-4(25-28)$
$=2(-39)-5(-30)-4(-3)$
$=-78+150+12$
$=84$ cube Unit

$\bar{a}(\bar{b} \times \bar{c})=[\bar{a} \bar{b} \bar{c}]=\left[\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right]$
$\therefore \bar{a}(\bar{b} \times \bar{c})=\left|\begin{array}{ccc}3 & -2 & 7 \\ 5 & 1 & -2 \\ 1 & 1 & -1\end{array}\right|$
$=3(-1+2)+2(-5+2)+7(5-1)$
$=3-6+28$
$=25$

R is the point which divides the line segment joining the points PQ internally in the ratio 2:1.
$\bar{r}=\frac{2(\bar{q})+1(\bar{p})}{2+1}$
$=\frac{2(\hat{i}+4 \hat{j}-2 \widehat{k})+1(\hat{i}-2 \hat{j}+\widehat{k})}{3}$
$=\frac{3 \hat{i}+6 \hat{j}-3 \widehat{k}}{3}$
$\bar{r}=\hat{i}+2 \hat{j}-\widehat{k}$
The position vector of point $R$ is $\hat{i}+2 \hat{j}-\widehat{k}$

Given that $\vec{a}=\hat{i}+2 \hat{j}, \vec{b}=-2 \hat{i}+\hat{j}, \vec{c}=4 \hat{i}+3 \hat{j}$
We need to find $x$ and $y$ such that $\vec{c}=x \vec{a}+y \vec{b}$
Substituting the values of $a$, $b$ and $c$, in $\vec{c}=x \vec{a}+y \vec{b}$, we have,
$4 \hat{i}+3 \hat{j}=x(\hat{i}+2 \hat{j})+y(-2 \hat{i}+\hat{j})$
$4 \hat{i}+3 \hat{j}=(x-2 y) \hat{i}+(2 x+y) \hat{j}$
Comparing the coefficients of i and j on both the sides, we have,
x-2y= 4
and
2x + y = 3
Solving the above simultaneous equations, we have,
x = 2 and y = -1

$2 \bar{a}+3 \bar{b}-5 \bar{c}=0$
$5 \bar{c}=3 \bar{b}+2 \bar{a}$
$\bar{c}=\frac{3 \bar{b}+2 \bar{a}}{5}$
$\bar{c}=\frac{3 \bar{b}+2 \bar{a}}{3+2}$
∴C divides seg AB internally in the ratio 3 : 2

Let $\overline{ a }=-3 \hat{ i }+4 \hat{ j }-2 \hat{ k }, \overline{ b }=\hat{ i }+2 \hat{ k }, \overline{ c }=\hat{ i }-p \hat{ j }$
Then $\bar{a}, \bar{b}, \bar{c}$ are coplanar.
$\therefore[\overline{ a } \overline{ b } \overline{ c }]=0$
$\therefore\left|\begin{array}{ccc}-3 & 4 & -2 \\ 1 & 0 & 2 \\ 1 & - p & 0\end{array}\right|=0$
∴ −3(0 + 2p) − 4(0 − 2) − 2(− p − 0) = 0
∴ − 6p + 8 + 2p = 0
∴ − 4p = − 8
∴ p = 2

$\begin{aligned} & =\left(2 \cos ^2 \alpha-1\right)+\left(2 \cos ^2 \beta-1\right)+\left(2 \cos ^2 \gamma-1\right) \\ & =2\left(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma\right)-3 \\ & =2(1)-3 \quad\left[\because \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\right] \\ & =-1 \\ & \therefore \cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=-1 \\ & \therefore \cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma+1=0\end{aligned}$

Given : $3 \bar{a}+5 \bar{b}-8 \bar{c}=0$
$
\begin{aligned}
& 3 \bar{a}=8 \bar{c}-5 \bar{b} \\
& \bar{a}=\frac{8 \bar{c}-5 \bar{b}}{3} \\
& \bar{a}=\frac{8 \bar{c}-5 \bar{b}}{8-5} \quad[\because 3=8-5]
\end{aligned}
$
$A(\bar{a})$ divides $BC$ externally in the ratio $8: 5$.

$\begin{aligned} & \bar{a} \cdot(\bar{b} \times \bar{c})=\left|\begin{array}{ccc}3 & -1 & 4 \\ 2 & 3 & -1 \\ -5 & 2 & 3\end{array}\right| \\ & =3(9+2)+1(6-5)+4(4+15) \\ & =33+1+76 \\ & =110\end{aligned}$

Let $\bar{a}, \bar{b}, \bar{c}$ and $\bar{d}$ be the position vectors of vertices $A , B , C , D$ respectively

∴From (i) and (ii), we get
$\bar{e}=\bar{f}$
The mid point of the diagonals AC and BD is same
∴ The diagonals AC and BD bisect each other.
$\therefore$ The $\square ABCD$ is a Parallelogram.

(C) 5/2
Let $\bar{a}=2 \hat{i}-q \hat{j}+3 \widehat{k}$ and $\bar{b}=4 \hat{i}-5 \hat{j}+6 \widehat{k}$
Since, $\bar{a}$ and $\bar{b}$ are collinear.
$\therefore$ there exists a scalar $t$ such that $\bar{b}=t \bar{a}$.
$\therefore 4 \hat{i}-5 \hat{j}+6 \widehat{k}=t(2 \hat{i}-q \hat{j}+3 \widehat{k})=2 t \hat{i}-q t \hat{j}+3 t \widehat{k}$
∴ By equality of vectors, we get
4 = 2t, - 5 = - qt, 6 = 3t
∵ 4 = 2t and 6 = 3t  ∴t = 2
- 5 =- q(2)
–5 = – 2q 
∴ 5 = 2q
q = 5/2