Answer

Verified

449.4k+ views

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Change the following sentences into negative and interrogative class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE