Question

if u = x + y + z, v = x2 + y2 + z2, w = yz +Zx + xy,
then show that grad u, grad v and grad w are
coplanar vector.​

Answer 1

Answer:

The required answer is $0$.

Step-by-step explanation:

• Three vectors are coplanar if their scalar triple product equals zero in a three-dimensional space.
• Three vectors in a three-dimensional space are coplanar if they are linearly independent of one another.
• In the situation of $n$ vectors, all vectors are coplanar if there are no more than two linearly independent vectors.

Given: $u = x + y + z, v = x2 + y2 + z2, w = yz +Zx + xy$

To find: grad $u$, grad $v$ and grad $w$ are coplanar vector.​

Solution:

The question based only on calculation:

$&\nabla=\frac{\partial}{\partial x} \hat{i}+\frac{\partial}{\partial y} \hat{\jmath}+\frac{\partial}{\partial z} \hat{k}$

$&\nabla u=\frac{\partial u}{\partial x} \hat{i}+\frac{\partial u}{\partial y} \hat{j}+\frac{\partial u}{\partial z} \hat{k}$

$&\nabla u=(1) \hat{i}+(1) \hat{j}+(1) \hat{k}$.......(1).

Calculating the scalar triple product is necessary to determine whether or not the three vectors $x, y,$ and $z$ are coplanar:

$\nabla V=& \frac{\partial}{\partial x}\left(x^{2}+y^{2}+z^{2}\right) \hat{i}+\frac{\partial}{\partial y}\left(x^{2}+y^{2}+z^{2}\right) \hat{j}+\frac{\partial}{\partial z}\left(x^{2}+y^{2}+z^{2}\right) \hat{k}$

$&\nabla v=(2 x) \hat{i}+2 y \hat{\jmath}+2 z \hat{k}$.....(2)

$\nabla \omega&=\frac{\partial}{\partial x}\{y z+z x+x y\} \hat{i}+\frac{\partial}{\partial y}\{y z+z x+x y) \hat{\jmath}+\frac{\partial}{\partial z}(y z+z x+x y) \hat{k}$

$\nabla \omega=(z+y) \hat{i}+(z+x) \hat{j}+(x+y) \hat{k}$.........(3)

three vectors are co-planar.

$&\Rightarrow \quad\left[$$\vec{a} & \vec{b} & \vec{c}]$$\end{array}\right]=0 \cdot \Rightarrow\left[\begi$$\nabla u & \nabla v & \nabla \omega]$

$\end{array}\right] \\&=\left|\begin{array}{ccc}1 & 1 & 1 \\2 x & 2 y & 2 z \\z+y & z+x & x+y\end{array}\right|$

$\begin{array}{r}=2 y(x+y)-2 z(z+x)-1(2 x(x+y)-2 z(z+y)) \\+1(z x(z+x)-2 y(z+y)) \\=2 x y+2 y^{2}-2 z^{2}-2 x z-2 x^{2}-2 x y+2 z^{2}+2 y z \\+2 x z+2 x^{2}-2 y z-2 y^{2}\end{array}\\\\=0$

As we can see, the vectors $x, y$ and $z$ are coplanar since the scalar triple product equals zero.

proved: grad $u$, grad $v$ and grad $w$ are coplanar vector.​

#SPJ2

Answer 2

Answer:

The 3 vectors are coplanar, and this can be proven by calculating their scalar triple product to be 0.

Step-by-step explanation:

We have learned that in a 3-dimensional space, 3 vectors are said to be coplanar if their triple product in the scalar method is equal to zero.

They also need to be independent of each other in a linear manner.

Given -

$u = x + y + z\\v = x^2 + y^2 + z^2\\w = yz + zx + xy$

To find - proof that the 3 given vectors are coplanar

Solution -

We can represent any given vector as $\bar v = \frac{\delta}{\delta x} i + \frac{\delta}{\delta y} j + \frac{\delta}{\delta z} k$.

$v_u = \frac{\delta}{\delta x} i + \frac{\delta}{\delta y} j + \frac{\delta}{\delta z} k\\\\v_u = (1)i + (1)j + (1)k$

To show that the vectors are coplanar, we need to determine the scalar product of these vectors.

$v_y = 2xi + 2yj + 2zk$
$v_z = (z+y)i + (z+x)j + (x+y)k$

When we calculate the scalar triple product of these 3 vectors, we need to get the product to be zero.

$|abc| = 0 = |v_u \; v_w \; v_z|$

$\left[\begin{array}{ccc}1&1&1\\2x&2y&2z\\z+y&z+x&x+y\end{array}\right]$

$= 2y(x+y) - 2z(z+x) - 1[2x(x+y) - 2z(z+y)] + 1[zx(z+x) - 2y(z+y)]\\= 2xy + 2y^2 - 2z^2 - 2xz - 3x^2 - 2xy + 2z^2 + 2yz + 2xz + 2x^2 - 2yz - 2y^2\\= 0$

Thus, since the scalar triple product of these vectors is 0, we can conclude that they are coplanar vectors.

#SPJ1