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Question: We have given vectors a and b by \[a = 3i - 2j + k\] and \[b = - 2i + 2j + 4k\] . Find the magnitude...

We have given vectors a and b by a=3i−2j+ka = 3i - 2j + k and b=−2i+2j+4kb = - 2i + 2j + 4k . Find the magnitude of a and b and the angle between them?

Explanation

Solution

We will find the magnitude of the vector using the magnitude formula. Then we will find the angle between the vectors using the dot product formula. The formula is A.B=∣A∣∣B∣cosθA.B = |A||B|cos\theta and then find the value of cos and then take its inverse.

Complete step-by-step answer:
We have given a=3i−2j+ka = 3i - 2j + k and b=−2i+2j+4kb = - 2i + 2j + 4k
We have to first find the magnitude of the vector a and b
We know that the magnitude of a vector having origin as centre is defined by
⇒∣v∣=x2+y2+z2\Rightarrow |v| = \sqrt {{x^2} + {y^2} + {z^2}} where x, y, z are the components the vector
So, the magnitude of vector a=3i−2j+ka = 3i - 2j + k
⇒∣v∣=x2+y2+z2\Rightarrow |v| = \sqrt {{x^2} + {y^2} + {z^2}}
⇒∣a∣=32+(−2)2+12\Rightarrow |a| = \sqrt {{3^2} + {{( - 2)}^2} + {1^2}}
⇒∣a∣=9+4+1\Rightarrow |a| = \sqrt {9 + 4 + 1}
{119.11^ \circ }$$$ \Rightarrow |a| = \sqrt {14} $$ --------- (1) Similarly, we will find the magnitude of vector $$b = - 2i + 2j + 4k$$ $$ \Rightarrow |v| = \sqrt {{x^2} + {y^2} + {z^2}} $$ $$ \Rightarrow |b| = \sqrt {{{\left( { - 2} \right)}^2} + {2^2} + {4^2}} $$ $$ \Rightarrow |b| = \sqrt {4 + 4 + 16} $$ $$ \Rightarrow |b| = \sqrt {24} $$ $$ \Rightarrow |b| = 2\sqrt 6 $$ -------(2) Now, we will find the angle between a and b vector We know that angle between two vector is find by using the dot product formula We will find the dot product of $$a = 3i - 2j + k$$ and $$b = - 2i + 2j + 4k$$ $$ = a \cdot b$$ $$ = (3 - 2 + 1)( - 2 + 2 + 4)$$ $$ = - 6$$ -------- (3) We know that $$A.B = |A||B|cos\theta $$ ,where θ is the angle between the vectors $$ \Rightarrow a.b = |a||b|cos\theta $$ We will put the magnitude of a and b from equation 1 and 2, value of dot product from equation 3 $$ \Rightarrow - 6 = \sqrt {14} \cdot 2\sqrt 6 \cdot cos(\theta )$$ We have to find the angle, so we take cos to left side $$ \Rightarrow cos(\theta ) = \dfrac{{ - 6}}{{\sqrt {14} .2\sqrt 6 }}$$ $$ \Rightarrow cos(\theta ) = \dfrac{{ - 3}}{{\sqrt {84} }}$$ We will find the value of angle by taking the inverse on both side $$ \Rightarrow {\cos ^{ - 1}}(\cos \theta ) = {\cos ^{ - 1}}\left( {\dfrac{{ - 3}}{{\sqrt {84} }}} \right)$$ We know that $${\cos ^{ - 1}}(\cos \theta ) = \theta $$ We will the value using scientific calculator $$ \Rightarrow \theta = {109.11^ \circ }$$ Hence, the magnitude of $$a = 3i - 2j + k$$ and $$b = - 2i + 2j + 4k$$ is \sqrt {14} andand2\sqrt 6 .theanglebetweenaandbis. the angle between a and b is{119.11^ \circ }$

Note: We have to be familiar with the formula for finding the magnitude of a vector in both conditions when they are stated from origin or when they are started and ended at some point. We also have to be familiar with the what is dot product and use of its formula for finding the angle between two vectors.