Question

Show that vectors a = 2i + 5j -6k and
b=i+5/2j^-3k^ are parallel.​

Answer 1

Two vectors A and B are parallel if and only if they are scalar multiples of one another.

Here b=1/2a. Therefore theyare parallel

Answer 2

Answer and Explanation:

Given : Vectors $\vec{a}=2i+5j-6k$ and $\vec{b}=i+\frac{5}{2}j-3k$

To find : Show that vectors are parallel?

Solution :

When two vectors are parallel then $\vec{u}\cdot\vec{v}=|a||b|$

So we find the dot product of vectors,

$\vec{a}\cdot\vec{b}=(2i+5j-6k)\cdot(i+\frac{5}{2}j-3k)$

$\vec{a}\cdot\vec{b}=2\times 1+5\times\frac{5}{2}+(-6)\times (-3)$

$\vec{a}\cdot\vec{b}=2+\frac{25}{2}+18$

$\vec{a}\cdot\vec{b}=20+\frac{25}{2}$

$\vec{a}\cdot\vec{b}=\frac{40+25}{2}$

$\vec{a}\cdot\vec{b}=\frac{65}{2}$ ...(1)

Now, we find the magnitude,

$\vec{a}=2i+5j-6k$

$|\vec{a}|=\sqrt{2^2+5^2+(-6)^2}$

$|\vec{a}|=\sqrt{4+25+36}$

$|\vec{a}|=\sqrt{65}$

$\vec{b}=i+\frac{5}{2}j-3k$

$|\vec{b}|=\sqrt{1^2+(\frac{5}{2})^2+(-3)^2}$

$|\vec{b}|=\sqrt{1+\frac{25}{4}+9}$

$|\vec{b}|=\sqrt{\frac{4+25+36}{4}}$

$|\vec{b}|=\sqrt{\frac{65}{4}}$

$|\vec{b}|=\frac{\sqrt{65}}{2}$

So, $|a||b|=\sqrt{65}\times \frac{\sqrt{65}}{2}$

$|a||b|=\frac{65}{2}$ ...(2)

Since, (1) and (2) are equal so the vectors are parallel.