If a unit vector a makes angles 3 with i, 4 with j and an acute angle with k, then find and hence, the components of a.

Let a=xi+y j+zk be a unit vector. ......(i) |a|=1 x^2+y^2+z^2 =1 Squaring both sides, x^2+y^2+z^2=1 .....(ii) Given: Angle between vectors a and i=i+0j+0k is 3 . 3 = a.i |a |. |i | 1 2 = x(1)+y(0)+z(0) (1)(1) 1 2 =x ....(iii)Again, given Angel between vectors a and j=0i+j+0k is 4 . 4 = a. j |a|.|j| 1 2 = x(0)+y(1)+z(0) (1)(1) 1 2 =y ......(iv) Again, given Angel between vectors a and k=0i+0j+k is , where is acute angle. = a. k |a|.|k| = x(0)+y(0)+z(1) (1)(1) =z .......(v) Putting the values of x, y and z in eq. (ii),1 4 +1 2 + ^2 =1 ^2 =1-1 4 -1 2 ^2 =4-1-2 4 =1 4 = 1 2 Since is acute angle, therefore cos is positive and hence 1 2 = 3 = 3 From eq. (v), z= =1 2 Putting values of x, y and z in eq. (i), a=1 2 i+1 2 j+1 2 k Components of a are coefficients of i,j,k in a 1 2 ,1 2 ,1 2 and angle = 3

If a unit vector a makes angles 3 with i, 4 with j and an acute a…

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Question

If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}\ \text{with}\ \hat{i},\frac{\pi}{4}\ \text{with}\ \hat{j}$ and an acute angle $\theta$ with $\hat{k},$ then find $\theta$ and hence, the components of $\vec{a}.$

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Answer

$\text{Let}\ \ \ \hat{a}=\text{x}\hat{i}+\text{y }\hat{j}+\text{z}\hat{k}\ \text{be a unit vector.}\ \ \ \ ......\text{(i)}$ $\Rightarrow\ \ |\hat{a}|=1\ \ \Rightarrow\ \ \sqrt{\text{x}^2+\text{y}^2+\text{z}^2}=1$ $\text{Squaring both sides,}\ \ \ \ \ \text{x}^2+\text{y}^2+\text{z}^2=1\ \ \ .....\text{(ii)}$ $\text{Given:}\ \ \text{Angle between vectors}\ \hat{a}\ \text{and}\ \hat{i}=\hat{i}+0\hat{j}+0\hat{k}\ \text{is}\ \frac{\pi}{3}.$ $\therefore\ \cos\frac{\pi}{3}=\frac{\hat{a}.\hat{i}}{\big|\hat{a}\big|.\big|\hat{i}\big|}\ $ $\Rightarrow\ \ \frac{1}{2}=\frac{\text{x}(1)+\text{y}(0)+\text{z}(0)}{(1)(1)}$ $\Rightarrow\ \ \ \frac{1}{2}=\text{x}\ \ \ ....\text{(iii)}$Again, given $\text{Angel between vectors}\ \hat{a}\ \text{and}\ \hat{j}=0\hat{i}+\hat{j}+0\hat{k}\ \text{is}\ \frac{\pi}{4}.$
$\therefore\ \cos\frac{\pi}{4}=\frac{\hat{a}.{\hat{j}}}{|\hat{a}|.|\hat{j}|}\ \Rightarrow\ \frac{1}{\sqrt{2}}=\frac{\text{x}(0)+\text{y}(1)+\text{z}(0)}{(1)(1)}$
$\Rightarrow\ \frac{1}{\sqrt{2}}=\text{y}\ \ ......\text{(iv)}$
Again, given $\text{Angel between vectors}\ \hat{a}\ \text{and}\ \hat{k}=0\hat{i}+0\hat{j}+\hat{k}\ \text{is}\ \theta,\ \text{where}\ \theta\ \text{is acute angle.}$
$\therefore\ \cos\theta=\frac{\hat{a}.{\hat{k}}}{|\hat{a}|.|\hat{k}|}\ $ $\Rightarrow\ \ \cos\theta=\frac{\text{x}(0)+\text{y}(0)+\text{z}(1)}{(1)(1)}$ $\Rightarrow\ \ \cos\theta=\text{z}\ \ \ .......\text{(v)}$ Putting the values of x, y and z in eq. (ii),$\frac{1}{4}+\frac{1}{2}+\cos^2\theta=1$ $\ \Rightarrow\ \ \cos^2\theta=1-\frac{1}{4}-\frac{1}{2}$
$\Rightarrow\ \ \ \ \ \ \ \cos^2\theta=\frac{4-1-2}{4}=\frac{1}{4}$ $\ \Rightarrow\ \ \cos\theta=\pm\frac{1}{2}$ Since $\theta$ is acute angle, therefore cos $\theta$ is positive and hence $\frac{1}{2}=\cos\frac{\pi}{3}\ \Rightarrow\ \ \theta=\frac{\pi}{3}$ From eq. (v), $\ \ \text{z}=\cos\theta=\frac{1}{2}$ Putting values of x, y and z in eq. (i), $\ \ \hat{a}=\frac{1}{2}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k}$$\therefore\ \ \text{Components of}\ \hat{a}\ \text{are coefficients of}\ \hat{i},\hat{j},\hat{k}\ \text{in}\ \hat{a}$
$\Rightarrow\ \ \frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2}\ \text{and angle}\ \theta=\frac{\pi}{3}$