# Let u, v and w be vectors such that $u+v+w=0$. If $\left| u \right|=3$, $\left| v \right|=4$ and $\left| w \right|=5$ then $u.v+v.w+w.u$ is equal to

(a) 0

(b) -25

(c) 25

(d) 50

(e) 47

Last updated date: 20th Mar 2023

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Answer

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Hint: We can take the square on both the sides of the equation, $u+v+w=0$. Then, we can use ${{\left( u+v+w \right)}^{2}}={{u}^{2}}+{{v}^{2}}+{{w}^{2}}+2\left( u.v+v.w+w.u \right)$, where we can write ${{u}^{2}}$ as ${{\left| u \right|}^{2}}$, ${{v}^{2}}$ as ${{\left| v \right|}^{2}}$ and ${{w}^{2}}$ as ${{\left| w \right|}^{2}}$. We can then substitute the values of $\left| u \right|=3,\left| v \right|=4,$ and $\left| w \right|=5$.

Complete step-by-step answer:

Before proceeding with the question, we must know the formula that will be required to solve this question. We must be very clear about the rules of dot product of vectors.

When we have a question involving the product of vectors, then we have to use the concept of dot products and apply the same.

Here in question, it is given that u + v + w = 0. But, we know that u, v and w are vectors, so their sum can be 0 only when all these three vectors are linearly dependent, or in the same plane.

In this question, it has been given that the sum of vectors $u+v+w=0$ and $\left| u \right|=3$, $\left| v \right|=4$ and $\left| w \right|=5$ and then we have been asked to find the value of $u.v+v.w+w.u$.

We know that, ${{\left( u+v+w \right)}^{2}}={{u}^{2}}+{{v}^{2}}+{{w}^{2}}+2\left( uv+vw+wu \right)$.

But in vector algebra, ${{\left( u+v+w \right)}^{2}}={{u}^{2}}+{{v}^{2}}+{{w}^{2}}+2\left( u.v+v.w+w.u \right)$ means that we have to replace $uv+vw+wu$ by dot product $u.v+v.w+w.u$.

Let us number the given equation as below,

$u+v+w=0.....(i)$

Now, squaring both sides of equation (i) we get,

$\Rightarrow {{\left( u+v+w \right)}^{2}}=0$

$\Rightarrow \left( {{u}^{2}}+{{v}^{2}}+{{w}^{2}}+2\left( u.v+v.w+w.u \right) \right)=0$

Taking the term involving dot products on the right-hand-side, we get,

$\Rightarrow \left( {{u}^{2}}+{{v}^{2}}+{{w}^{2}} \right)=-2\left( u.v+v.w+w.u \right)$

Now, replacing ${{u}^{2}}$ as ${{\left| u \right|}^{2}}$, ${{v}^{2}}$ as ${{\left| v \right|}^{2}}$ and ${{w}^{2}}$ as ${{\left| w \right|}^{2}}$, we get,$\Rightarrow \left( {{\left| u \right|}^{2}}+{{\left| v \right|}^{2}}+{{\left| w \right|}^{2}} \right)=-2\left( u.v+v.w+w.u \right).....(ii)$

We know that $\left| u \right|=3$, $\left| v \right|=4$ and $\left| w \right|=5$. Therefore, we have to take the square of the values of $\left| u \right|=3$, $\left| v \right|=4$ and $\left| w \right|=5$ to get the value of ${{\left| u \right|}^{2}}$, ${{\left| v \right|}^{2}}$ and ${{\left| w \right|}^{2}}$.

$\therefore {{\left| u \right|}^{2}}=9,{{\left| v \right|}^{2}}=16,$ and ${{\left| w \right|}^{2}}=25$

Substituting the value of ${{\left| u \right|}^{2}}=9,{{\left| v \right|}^{2}}=16$, and ${{\left| w \right|}^{2}}=25$ in equation (ii) we get,

$\Rightarrow \left( 9+16+25 \right)=-2\left( u.v+v.w+w.u \right)$

Taking (-2) on left-hand-side we get,

$\Rightarrow \dfrac{\left( 50 \right)}{-2}=\left( u.v+v.w+w.u \right)$

After dividing 50 with 2 we get,

$\Rightarrow -25=\left( u.v+v.w+w.u \right)$

Therefore, we have obtained the value of $u.v+v.w+w.u$ as -25.

Hence, the correct answer is option (b).

Note: The simplification and operations on vectors is not the same way that we do in case of algebraic operations. So, one must be very careful about using the rules of vector addition and squaring vectors. The dot product of vectors must be used properly. When there is modulus given in the question, one must be careful about the signs.

Complete step-by-step answer:

Before proceeding with the question, we must know the formula that will be required to solve this question. We must be very clear about the rules of dot product of vectors.

When we have a question involving the product of vectors, then we have to use the concept of dot products and apply the same.

Here in question, it is given that u + v + w = 0. But, we know that u, v and w are vectors, so their sum can be 0 only when all these three vectors are linearly dependent, or in the same plane.

In this question, it has been given that the sum of vectors $u+v+w=0$ and $\left| u \right|=3$, $\left| v \right|=4$ and $\left| w \right|=5$ and then we have been asked to find the value of $u.v+v.w+w.u$.

We know that, ${{\left( u+v+w \right)}^{2}}={{u}^{2}}+{{v}^{2}}+{{w}^{2}}+2\left( uv+vw+wu \right)$.

But in vector algebra, ${{\left( u+v+w \right)}^{2}}={{u}^{2}}+{{v}^{2}}+{{w}^{2}}+2\left( u.v+v.w+w.u \right)$ means that we have to replace $uv+vw+wu$ by dot product $u.v+v.w+w.u$.

Let us number the given equation as below,

$u+v+w=0.....(i)$

Now, squaring both sides of equation (i) we get,

$\Rightarrow {{\left( u+v+w \right)}^{2}}=0$

$\Rightarrow \left( {{u}^{2}}+{{v}^{2}}+{{w}^{2}}+2\left( u.v+v.w+w.u \right) \right)=0$

Taking the term involving dot products on the right-hand-side, we get,

$\Rightarrow \left( {{u}^{2}}+{{v}^{2}}+{{w}^{2}} \right)=-2\left( u.v+v.w+w.u \right)$

Now, replacing ${{u}^{2}}$ as ${{\left| u \right|}^{2}}$, ${{v}^{2}}$ as ${{\left| v \right|}^{2}}$ and ${{w}^{2}}$ as ${{\left| w \right|}^{2}}$, we get,$\Rightarrow \left( {{\left| u \right|}^{2}}+{{\left| v \right|}^{2}}+{{\left| w \right|}^{2}} \right)=-2\left( u.v+v.w+w.u \right).....(ii)$

We know that $\left| u \right|=3$, $\left| v \right|=4$ and $\left| w \right|=5$. Therefore, we have to take the square of the values of $\left| u \right|=3$, $\left| v \right|=4$ and $\left| w \right|=5$ to get the value of ${{\left| u \right|}^{2}}$, ${{\left| v \right|}^{2}}$ and ${{\left| w \right|}^{2}}$.

$\therefore {{\left| u \right|}^{2}}=9,{{\left| v \right|}^{2}}=16,$ and ${{\left| w \right|}^{2}}=25$

Substituting the value of ${{\left| u \right|}^{2}}=9,{{\left| v \right|}^{2}}=16$, and ${{\left| w \right|}^{2}}=25$ in equation (ii) we get,

$\Rightarrow \left( 9+16+25 \right)=-2\left( u.v+v.w+w.u \right)$

Taking (-2) on left-hand-side we get,

$\Rightarrow \dfrac{\left( 50 \right)}{-2}=\left( u.v+v.w+w.u \right)$

After dividing 50 with 2 we get,

$\Rightarrow -25=\left( u.v+v.w+w.u \right)$

Therefore, we have obtained the value of $u.v+v.w+w.u$ as -25.

Hence, the correct answer is option (b).

Note: The simplification and operations on vectors is not the same way that we do in case of algebraic operations. So, one must be very careful about using the rules of vector addition and squaring vectors. The dot product of vectors must be used properly. When there is modulus given in the question, one must be careful about the signs.

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