Question

Last one for today!!!

Topic : Vectors

Give a detailed explanation please❤!​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\vec{a} = \dfrac{\hat{i} - 2\hat{j} + 2\hat{k}}{3}$

$\rm :\longmapsto\:\vec{b} = \dfrac{ - 4\hat{i}- 3\hat{k}}{5}$

$\rm :\longmapsto\:\vec{c} = \hat{j}$

Now,

$\rm :\longmapsto\: |\vec{a}| = \sqrt{ \dfrac{1}{9} + \dfrac{4}{9} + \dfrac{4}{9}} = \sqrt{\dfrac{1 + 4 + 4}{9}} = 1$

$\rm :\longmapsto\: |\vec{b}| = \sqrt{ \dfrac{9}{25} + \dfrac{16}{25}} = \sqrt{ \dfrac{25}{25} } = 1$

$\rm :\longmapsto\: |\vec{c}| = 1$

Now,

Let us assume that

$\red{\rm :\longmapsto\:\vec{d} = x\hat{i} + y\hat{j} + z\hat{k} \: which \: makes \: equal \: angle}$

$\red{\rm :\longmapsto\: \alpha, \beta, \gamma \: with\vec{a}, \: \vec{b}, \: \vec{c} \: respectively}$

So, According to statement,

$\rm :\longmapsto\:cos \alpha = cos \beta = cos \gamma$

$\rm :\longmapsto\:\dfrac{\vec{d}.\vec{a}}{ |\vec{d}| |\vec{a}| } = \dfrac{\vec{d}.\vec{b}}{ |\vec{d}| |\vec{b}| } = \dfrac{\vec{d}.\vec{c}}{ |\vec{d}| |\vec{c}| }$

$\rm :\longmapsto\:\dfrac{x - 2y + 2z}{3 \sqrt{ {x}^{2} + {y}^{2} + {z}^{2} } } = \dfrac{ - 4x - 3z}{5\sqrt{ {x}^{2} + {y}^{2} + {z}^{2} }} = \dfrac{y}{\sqrt{ {x}^{2} + {y}^{2} + {z}^{2} }}$

$\rm :\longmapsto\:\dfrac{x - 2y + 2z}{3 } = \dfrac{ - 4x - 3z}{5} = y$

Taking first and third member, we get

$\rm :\longmapsto\:\dfrac{x - 2y + 2z}{3 }= y$

$\rm :\longmapsto\:x - 2y + 2z = 3y$

$\rm :\longmapsto\:x - 5y + 2z = 0 - - - (1)$

Now,

Taking second and third member, we get

$\rm :\longmapsto\: \dfrac{ - 4x - 3z}{5} = y$

$\rm :\longmapsto\: - 4x - 3z = 5y$

$\rm :\longmapsto\: 4x + 5y + 3z = 0 - - - (2)$

Now, Solving equation (1) and (2) using cross multiplication method, we get

$\rm :\longmapsto\:\dfrac{x}{ - 15 - 10} = \dfrac{y}{8 - 3} = \dfrac{z}{5 + 20}$

$\rm :\longmapsto\:\dfrac{x}{ - 25} = \dfrac{y}{5} = \dfrac{z}{25}$

$\rm :\longmapsto\:\dfrac{x}{ - 5} = \dfrac{y}{1} = \dfrac{z}{5}$

$\rm :\longmapsto\:\dfrac{x}{5} = \dfrac{y}{ - 1} = \dfrac{z}{ - 5}$

$\rm :\longmapsto\:x = 5k$

$\rm :\longmapsto\:y = - k$

$\rm :\longmapsto\:z = -5 k$

So,

$\red{\rm :\longmapsto\:\vec{d} = 5k\hat{i} -k \hat{j} - 5k\hat{k} }$

As,

$\rm :\longmapsto\: |\vec{d}| = 1$

$\rm :\longmapsto\: \sqrt{ 25 {k}^{2} + {k}^{2} + {25k}^{2} } = \sqrt{51}$

$\rm :\longmapsto\:51 {k}^{2} = 51$

$\rm :\longmapsto\: {k}^{2} = 1$

$\rm :\longmapsto\:k \: = \: \pm \: 1$

Thus,

$\red{\rm :\longmapsto\:\vec{d} = \pm \: ( 5\hat{i} -\hat{j} - 5\hat{k} )}$

So,

Required vector is

$\red{\rm :\longmapsto\:\vec{d} = 5\hat{i} - \hat{j} - 5\hat{k} }$

and

$\rm :\longmapsto\: |\vec{d}| = \sqrt{25 + 1 + 25} = \sqrt{51}$

Hence,

Option (a) is correct.