Winter 2016 – Q.7

7. a) Show that.
$latex \left(\overrightarrow a\times\overrightarrow d\right)\times\left(\overrightarrow c\times\overrightarrow d\right)+\left(\overrightarrow a\times\overrightarrow c\right)\times\left(\overrightarrow d\times\overrightarrow b\right)+\left(\overrightarrow a\times\overrightarrow d\right)\times\left(\overrightarrow b\times\overrightarrow c\right)$
is paralled to the vector $latex \overrightarrow a.$. [6M]

Consider L.H.S $latex =\left(\;\overline a\times\overline b\;\right)\times\left(\;\overline c\times\overline d\;\right)+\left(\;\overline a\times\overline c\;\right)\times\left(\;\overline d\times\overline b\;\right)+\left(\;\overline a\times\overline d\;\right)\times\left(\;\overline b\times\overline c\right)$

$latex =\overline c\;\left[\overline a\cdot\left(\;\overline b\times\overline d\;\right)\right]-\overline d\;\left[\overline a\cdot\left(\;\overline b\times\overline c\;\right)\right]+\overline b\;\left[\overline c\cdot\left(\;\overline a\times\overline d\;\right)\right]-\overline d\;\left[\overline c\cdot\left(\;\overline a\times\overline b\;\right)\right]+\overline c\;\left[\overline b\cdot\left(\;\overline a\times\overline d\;\right)\right]-\overline b\;\left[\overline c\cdot\left(\;\overline a\times\overline d\;\right)\right]$

$latex =-\overline d\;\left[\;\overline a\cdot\left(\;\overline b\times\overline c\;\right)\right]-\overline d\;\left[\;\overline a\cdot\left(\;\overline b\times\overline c\;\right)\right]$

$latex \begin{array}{l}\because\left[\;\overline a\cdot\overline b\times\overline c=\overline b\cdot\overline c\times\overline a=\overline c\cdot\overline a\times\overline b\;and\;\overline d\cdot\overline b\times\overline c=-\left(\;\overline a\cdot\overline c\times\overline b\;\right)\right]\\\\=-2\;\overline d\;\left[\;\overline a\cdot\left(\;\overline b\times\overline c\;\right)\right]\\\\=u\cdot\overline d\end{array}$

Where $latex u=-2\left[\;\overline a\cdot\left(\;\overline b\times\overline c\right)\right]$ a scalar

Now L.H.S is scalar multiple of $latex \overline d$ and hence parallel to $latex \overline d$

$latex \therefore\left(\;\overline a\times\overline b\;\right)\times\left(\;\overline c\times\overline d\;\right)+\left(\;\overline a\times\overline c\;\right)\times\left(\;\overline d\times\overline b\;\right)+\left(\;\overline a\times\overline d\;\right)\times\left(\;\overline b\times\overline c\right)=-2\left[\;\overline a\cdot\;\overline b\times\overline c\right]\;\overline d$

b) Find the directional derivative of $latex \phi\;\left(x,\;y,\;z\right)=x^2-2y^2+4z^2$ at the point (1, 1, -1) in the direction 2i + j – k. In what direction will the directional derivative be maximum and what is its magnitude? [6M]

Given that $latex \phi\;\left(x,\;y,\;z\right)=x^2-2y^2+4z^2$

$latex \therefore\nabla\phi=\left(i\frac\partial{\partial x}+j\frac\partial{\partial y}+k\frac\partial{\partial z}\right)\;\left(x^2-2y^2+4z^2\right)$

$latex \begin{array}{l}=i\frac\partial{\partial x}x^2+j\frac\partial{\partial y}\left(-2y^2\right)+k\frac\partial{\partial z}\;\left(4z^2\right)\\\\=2xi-4yj+8zk\\\\\therefore{\left(\nabla\phi\right)}_{\left(1,\;1,\;-1\right)}=2i-4j-8k\\\\Let\;\overline a=2i+j-k\end{array}$

∴ unit vector in the direction of $latex \overline a$ is

$latex \begin{array}{l}\widehat a=\frac{2i+j-k}{\sqrt{2^2+1^2+\left(-1\right)^2}}\\\\\;\;\;=\frac1{\sqrt{4+!+1}}\left(2i+j-k\right)\\\\\widehat a=\frac{2i+j-k}{\sqrt6}\end{array}$

∴ directional derivatives of $latex \phi=\nabla\phi\cdot\widehat a$

$latex \begin{array}{l}=\left(2i-4j-8k\right)\cdot\frac1{\sqrt6}\left(2i+j-k\right)\\\\=\frac1{\sqrt6}\left(4-4+8\right)\\\\=\frac8{\sqrt6}\end{array}$

∴ directional derivative is maximum in the direction of $latex \nabla\phi\;i.e.\;2i-4j-8k$

its magnitude $latex =\left|\nabla\phi\right|=\sqrt{2^2+\left(-4\right)^2+\left(-8\right)^2}$

$latex \begin{array}{l}=\sqrt{4+16+64}\\\\=\sqrt{84}\\\\=\sqrt{21\times4}\\\\\left|\nabla\phi\right|=2\sqrt{21}\end{array}$

c) Prove that $latex \overline A=\left(6xy+z^3\right)i+\left(3x^2-z\right)j+\left(3xz^2-y\right)k$ is irrotational. Find the scaler potential $latex \phi$ such that $latex A=\triangle\phi.$ [6M]

We have,

$latex \overline A=\left(6xy+z^3\right)i+\left(3x^2-z\right)j+\left(3xz^2-y\right)k$

$latex \nabla\times A=\begin{vmatrix}i&j&k\\\frac\partial{\partial x}&\frac\partial{\partial y}&\frac\partial{\partial z}\\6xy+z^3&3x^2-z&3xz^2-y\end{vmatrix}$

$latex \nabla\times A=i\left[\frac\partial{\partial y}\left(3xz^2-y\right)-\frac\partial{\partial z}\left(3x^2+z\right)\right]-j\left[\frac\partial{\partial x}\left(3xz^2-y\right)-\frac\partial{\partial z}\left(6xy+z^3\right)\right]+k\left[\frac\partial{\partial x}\left(3x^2-z\right)-\frac\partial{\partial y}\left(6xy+z^3\right)\right]$

$latex \begin{array}{l}=i\left[\left(-1\right)-\left(-1\right)\right]-j\left[3z^2-3z^2\right]+k\left(6x-6x\right)\\\\\nabla\times\overline A=0\end{array}$

∴ $latex \overline A$ is irrotational

Let $latex \phi$ be the scalar potential of $latex \overline A$ such that

$latex \overline A=\nabla\phi$

i.e. $latex \left(6xy+z^3\right)i+\left(3x^2-z\right)j+\left(3xz^2-y\right)k$

$latex =i\frac{\partial\phi}{\partial x}+j\frac{\partial\phi}{\partial y}+k\frac{\partial\phi}{\partial z}$

$latex \therefore\frac{\partial\phi}{\partial x}=6xy+z^3,\;\frac{\partial\phi}{\partial y}=3x^2-z,\;\frac{\partial\phi}{\partial z}=3xz^2-y$

on integrating

$latex \begin{array}{l}\phi=3x^2y+z^3x+\phi_1\;\left(y,\;z\right)—–\left(i\right)\\\\\phi=3x^2y-zy+\phi_2\;\left(x,\;z\right)—–\left(ii\right)\\\\\phi=xz^3-yz+\phi_3\;\left(x,\;y\right)—–\left(iii\right)\end{array}$

By compairing (i) (ii) and (iii)

$latex \begin{array}{l}\phi_1\;\left(y,\;z\right)=terms\;in\;\phi\;independent\;of\;x=-zy\\\\\phi_2\;\left(x,\;z\right)=terms\;in\;\phi\;independent\;of\;y=z^3x\\\\\phi_3\;\left(x,\;y\right)=terms\;in\;\phi\;independent\;of\;z=3x^2y\end{array}$

$latex \phi=3x^2y+z^3x-zy$

$latex \therefore\phi=3x^2y+z^3x-zy$