Show that vector a= 2i + 5j - 6k and vector b= i+5/2j-3k arr parallel
Answers
Answer:
Vectors \vec{a}=2i+5j-6k
a
=2i+5j−6k and \vec{b}=i+\frac{5}{2}j-3k
b
=i+
2
5
j−3k
To find : Show that vectors are parallel?
Solution :
When two vectors are parallel then \vec{u}\cdot\vec{v}=|a||b|
u
⋅
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)
a
⋅
b
=(2i+5j−6k)⋅(i+
2
5
j−3k)
\vec{a}\cdot\vec{b}=2\times 1+5\times\frac{5}{2}+(-6)\times (-3)
a
⋅
b
=2×1+5×
2
5
+(−6)×(−3)
\vec{a}\cdot\vec{b}=2+\frac{25}{2}+18
a
⋅
b
=2+
2
25
+18
\vec{a}\cdot\vec{b}=20+\frac{25}{2}
a
⋅
b
=20+
2
25
\vec{a}\cdot\vec{b}=\frac{40+25}{2}
a
⋅
b
=
2
40+25
\vec{a}\cdot\vec{b}=\frac{65}{2}
a
⋅
b
=
2
65
...(1)
Now, we find the magnitude,
\vec{a}=2i+5j-6k
a
=2i+5j−6k
|\vec{a}|=\sqrt{2^2+5^2+(-6)^2}∣
a
∣=
2
2
+5
2
+(−6)
2
|\vec{a}|=\sqrt{4+25+36}∣
a
∣=
4+25+36
|\vec{a}|=\sqrt{65}∣
a
∣=
65
\vec{b}=i+\frac{5}{2}j-3k
b
=i+
2
5
j−3k
|\vec{b}|=\sqrt{1^2+(\frac{5}{2})^2+(-3)^2}∣
b
∣=
1
2
+(
2
5
)
2
+(−3)
2
|\vec{b}|=\sqrt{1+\frac{25}{4}+9}∣
b
∣=
1+
4
25
+9
|\vec{b}|=\sqrt{\frac{4+25+36}{4}}∣
b
∣=
4
4+25+36
|\vec{b}|=\sqrt{\frac{65}{4}}∣
b
∣=
4
65
|\vec{b}|=\frac{\sqrt{65}}{2}∣
b
∣=
2
65
So, |a||b|=\sqrt{65}\times \frac{\sqrt{65}}{2}∣a∣∣b∣=
65
×
2
65
|a||b|=\frac{65}{2}∣a∣∣b∣=
2
65
...(2)
Since, (1) and (2) are equal so the vectors are parallel.