Question

find the value of X and y for which vectors A=6i^+xj^-2k^,B=5i^-6j^-yk^ are parallel​

Answer 1

Answer:

Explanation:

If vectors are parallel, then

a1/a2 = b1/b2 = c1/c2

6/5 = - x/6 = 2/y

x = - 36/5

2/y = 6/5

y = 5/3

Answer 2

Answer:

The value of x= -36/5 and y= 5/3.

Explanation:

If two vectors are parallel then then the cross product must of two vectors be 0.

A X B =0

Given:

$\vec{A}=6 \hat{i}+x \hat{\jmath}-2 \hat{k} \\ \vec{B}=5 \hat{\imath}-6 \hat{\jmath}-y \hat{k}^{}$

$\begin{aligned}&A \times B=\left|\begin{array}{ccc}i & j & k \\6 & x & -2 \\5 & -6 & -y\end{array}\right|=0 \\&i(-x y-12)-j(-6 y-(-10))+k(-36-5 x)=0\end{aligned}$

$Comparing LHS and RHS \begin{aligned}&-6y+10=0 \\&6y=10 \\y=\frac{10}{6}=\frac{5}{3}\end{aligned} Also, \begin{gathered}-36-5x=0 \\5x=-36 \\x=\frac{-36}{5} \\\end{gathered}$

The value of  $x=\frac{-36}{5} and y=\frac{5}{3}$.