Question

these are 3 seperate questions please answer abcd for each

5 31 and v= 2 3 -3 Calculate the dot product of u = 0-2 0 8 -8

Answer 1

1) dot product of two vectors can be found by

21 C2 = I1 * x2 + yi * Y2 + 21 * 22 Yi 21 Y2 22

so

$\large u\cdot v=\begin{bmatrix} -1\\3\\3 \end{bmatrix}\cdot \begin{bmatrix} 5\\2\\-3 \end{bmatrix}=-1*5+3*2+3*-3=-8$

so answer is -8

option d

2) by using given matrices ax+by=z can be written as

$\large a*\begin{bmatrix} 4\\1\\-2 \end{bmatrix}+b\begin{bmatrix} -1\\3\\3 \end{bmatrix}=\begin{bmatrix} -14\\3\\12 \end{bmatrix}$

then matrix form for this is

$\large \begin{bmatrix} 4&-1\\1&3\\-2&3 \end{bmatrix}\begin{bmatrix} a\\b \end{bmatrix}=\begin{bmatrix} -14\\3\\12 \end{bmatrix}$

augmented matrix for this is

4 -1|-14] 1 3 3 -23 12

convert this to row reduced echelon form , then last column gives result . on converting this becomes

$\large \begin{bmatrix} 1&0\\0&1\\0&0 \end{matrix}\begin{vmatrix} -3\\2\\0 \end{bmatrix}$

so solution is  a=-3 b=2

so answer is option a

3) equation for a plane passing through points  $\large (x_1,y_1,z_1)$ ,$\large (x_2,y_2,z_2)$ and $\large (x_3,y_3,z_3)$ is given by the following equation containing determinent .

$\large \begin{vmatrix} x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1 \end{vmatrix}=0$

using given 3 points equation becomes

$\large \begin{vmatrix} x-1&y-1&z-1\\3-1&-1-1&5-1\\4-1&2-1&-2-1 \end{vmatrix}=0$

$\large \begin{vmatrix} x-1&y-1&z-1\\2&-2&4\\3&1&-3 \end{vmatrix}=0$

now solve this

(x-1)[6-4] -(y-1)[-6-12]+(z-1)[2+6]=0

2(x-1)+18(y-1)+8(z-1)=0

2x+18y+8z= 28

divide both side by 2 to get

x+9y+4z=14

so answer is

x+9y+4z=14

option b