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Question: Let u and v are unit vectors and w is a vector such that \[u \times v + u = w\,and\,w \times u = v\]...

Let u and v are unit vectors and w is a vector such that u×v+u=w and w×u=vu \times v + u = w\,and\,w \times u = v, then find the value of [u⃗ v⃗ w⃗][\vec u\,\vec v\,\vec w] ?

Explanation

Solution

Here the given question is of vector algebra, where we need to solve according to the properties of vector algebra, in order to solve it we know that unit vector are the vector having a magnitude of one, and some properties like commutative and distributive property will also be used here.

Formulae Used:
⇒[u⃗ v⃗ w⃗]=u⃗.(v⃗×w⃗)\Rightarrow [\vec u\,\vec v\,\vec w] = \vec u.(\vec v \times \vec w)

Complete step by step answer:
The given question need to be solved with the help of rules related with the vector algebra in order to get the solution for this question, here first we need to expand the given conditions and then try to reach to the asked question, on solving we get:
Given that:

⇒u⃗×v⃗+u⃗=w⃗ and w⃗×u⃗=v⃗ ⇒(u⃗×v⃗+u⃗)×u⃗=w⃗×u⃗ ⇒(u⃗×v⃗)×u⃗+u⃗×u⃗=v⃗ (as w⃗×u⃗=v⃗) ⇒(u⃗.u⃗)v⃗−(v⃗.u⃗)u⃗+u⃗×u⃗=v⃗  \Rightarrow \vec u \times \vec v + \vec u = \vec w\,and\,\vec w \times \vec u = \vec v \\\ \Rightarrow \left( {\vec u \times \vec v + \vec u} \right) \times \vec u = \vec w \times \vec u \\\ \Rightarrow \left( {\vec u \times \vec v} \right) \times \vec u + \vec u \times \vec u = \vec v\,(as\,\vec w \times \vec u = \vec v) \\\ \Rightarrow \left( {\vec u.\vec u} \right)\vec v - \left( {\vec v.\vec u} \right)\vec u + \vec u \times \vec u = \vec v \\\

Using, u⃗.u⃗=1, and u⃗×u⃗=0\vec u.\vec u = 1,\,and\,\vec u \times \vec u = 0, since the dot product of the unit vector is one, and the cross product of the unit vector is zero.

⇒(1)v⃗−(v⃗.u⃗)u⃗+0=v⃗ ⇒v⃗−v⃗=(v⃗.u⃗)u⃗ ⇒(v⃗.u⃗)u⃗=0 ⇒u⃗.v⃗=0 (as u⃗≠0)..................................................(eq1) ⇒[u⃗ v⃗ w⃗]=u⃗.(v⃗×w⃗) ⇒given u⃗×v⃗+u⃗=w⃗ ⇒u⃗.(v⃗×(u⃗×v⃗+u⃗)) ⇒u⃗.(v⃗×(u⃗×v⃗)+v⃗×u⃗) ⇒u⃗.((v⃗.v⃗)u⃗−(v⃗.u⃗)v⃗+v⃗×u⃗) ⇒u⃗.(∣v⃗2∣u⃗−0+v⃗×u⃗) sin⁡ce (v⃗.u⃗=0,from eq1) ⇒∣v⃗2∣(u⃗.u⃗)+u⃗.(v⃗×u⃗)  \Rightarrow \left( 1 \right)\vec v - \left( {\vec v.\vec u} \right)\vec u + 0 = \vec v \\\ \Rightarrow \vec v - \vec v = \left( {\vec v.\vec u} \right)\vec u \\\ \Rightarrow \left( {\vec v.\vec u} \right)\vec u = 0 \\\ \Rightarrow \vec u.\vec v = 0\,(as\,\vec u \ne 0)..................................................(eq1) \\\ \Rightarrow [\vec u\,\vec v\,\vec w] = \vec u.(\vec v \times \vec w) \\\ \Rightarrow given\,\vec u \times \vec v + \vec u = \vec w \\\ \Rightarrow \vec u.(\vec v \times \left( {\vec u \times \vec v + \vec u} \right)) \\\ \Rightarrow \vec u.(\vec v \times (\vec u \times \vec v) + \vec v \times \vec u) \\\ \Rightarrow \vec u.((\vec v.\vec v)\vec u - (\vec v.\vec u)\vec v + \vec v \times \vec u) \\\ \Rightarrow \vec u.(|{{\vec v}^2}|\vec u - 0 + \vec v \times \vec u)\,\sin ce\,(\vec v.\vec u = 0,from\,eq1) \\\ \Rightarrow |{{\vec v}^2}|(\vec u.\vec u) + \vec u.(\vec v \times \vec u) \\\ ⇒∣v⃗2∣∣u⃗∣2+0 ⇒1 (sin⁡ce ∣v⃗∣,∣u⃗∣=1 unit vectors) ⇒∴[u⃗ v⃗ w⃗]=1  \Rightarrow |{{\vec v}^2}||\vec u{|^2} + 0 \\\ \Rightarrow 1\,(\sin ce\,|\vec v|,|\vec u| = 1\,unit\,vectors) \\\ \Rightarrow \therefore [\vec u\,\vec v\,\vec w] = 1 \\\

Here we get the answer for the question, the asked magnitude value is one.

Note: Here the given question is of vector algebra and vector algebra works on its properties, multiplication, division has its own rule, here there are two types of multiplication, one is dot product and the other one is cross product and they act separately.