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Question: The \(W = \left| u \right|*v + \left| v \right|*u\) where u, v and W are non-zero vectors. show that...

The W=∣u∣∗v+∣v∣∗uW = \left| u \right|*v + \left| v \right|*u where u, v and W are non-zero vectors. show that W bisects the angle between u and v?

Explanation

Solution

Here in this question, we are given three non-zero vectors and we are provided with a relation between the vectors, we have to prove that whether the third vector bisects the two vectors. For that we have to prove the angles are equal or not. Use the dot product formula for the two cases.

Complete step by step answer:
We are provided that u, v and W are non-zero vectors. Vectors are the quantities which have magnitude as well as direction. These are represented by arrows which are made at the top of the vectors represented as u⃗\vec u. Vectors are represented as i^,j^,k^\hat i,\hat j,\hat k which represents x, y and z axis respectively. From the relation given in the question, we can say that the W⃗\vec W lies in the plane of u⃗\vec u and v⃗\vec v. Let us consider θu{\theta _u} and θv{\theta _v} to be the angles the vector W⃗\vec W made with u⃗\vec u and v⃗\vec vrespectively. Them using the formula,
W⃗⋅u⃗=∣W⃗∣∣u⃗∣cos⁡θu\vec W \cdot \vec u = \left| {\vec W} \right|\left| {\vec u} \right|\cos {\theta _u}
Using the given relation,
W⃗⋅u⃗=(∣u⃗∣v⃗+∣v⃗∣u⃗)⋅u⃗\vec W \cdot \vec u = \left( {\left| {\vec u} \right|\vec v + \left| {\vec v} \right|\vec u} \right) \cdot \vec u
On solving,
∣W⃗∣∣u⃗∣cos⁡θu=∣u⃗∣(v⃗⋅u⃗+∣v⃗∣∣u⃗∣) ∣W∣cos⁡θu=(v⃗⋅u⃗+∣v⃗∣∣u⃗∣)  \left| {\vec W} \right|\left| {\vec u} \right|\cos {\theta _u} = \left| {\vec u} \right|\left( {\vec v \cdot \vec u + \left| {\vec v} \right|\left| {\vec u} \right|} \right) \\\ \left| W \right|\cos {\theta _u} = \left( {\vec v \cdot \vec u + \left| {\vec v} \right|\left| {\vec u} \right|} \right) \\\
Similarly,
∣W⃗∣cos⁡θv=u⃗⋅v⃗+∣u⃗∣∣v⃗∣\left| {\vec W} \right|\cos {\theta _v} = \vec u \cdot \vec v + \left| {\vec u} \right|\left| {\vec v} \right|
By interchanging u⃗\vec u with v⃗\vec v and vice-versa, we find that the above equations become equal.
∣W⃗∣cos⁡θu=∣W⃗∣cos⁡θv\left| {\vec W} \right|\cos {\theta _u} = \left| {\vec W} \right|\cos {\theta _v}
Which states that, θu=θv{\theta _u} = {\theta _v}

Hence, we prove that W⃗\vec W bisects the angle between u⃗\vec u and v⃗\vec v.

Note: The above question has a special case where the two vectors will not bisect that is when u⃗=−v⃗\vec u = - \vec v because it will make the W⃗=0\vec W = 0. Bisects means that the dividing the angle into two, which can be proven by proving the angles for the two is equal.