Question

A line makes an angle theta1, theta2, theta3 and theta 4 with diagonals of a cube.Show that cos^2theta1+cos^2theta2+cos^2theta3+cos^2theta4=4/3....
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Answer 1

$\ cos^2\Theta_1 + cos^2\Theta_2 + cos^2\Theta_3 + cos^2\Theta_4 = \frac {4}{3}$ Proved

Step-by-step explanation:

Refer Figure Attached to Understand Easily,

Take O, a corner of Cube OBLCMANP, as origin and OA, OB, OC, the three edges through it as the axes.

Let OA = OB = OC = a, then the coordinates of O, A, B, C are (0,0,0,), (a,0,0,), (0,a,0),(a,a,0) respectively.

OP, AL, BM, CN be the Diagonal of Cube (4 in Number)

• Direction cosines of OP are proportional to a-0, a-0, a-0 or a, a, a, i.e. 1, 1, 1
• Direction cosines of AL are proportional to 0-a, a-0, a-0 or -a, a, a, i.e. -1, 1, 1
• Direction cosines of BM are proportional to a-0, 0-a, a-0 or a, -a, a, i.e. 1, -1, 1
• Direction cosines of CN are proportional to a-0, a-0, 0-a or a, a, -a, i.e. 1, 1, -1

Hence,  

• Direction Cosines of OP are $\frac {1}{\sqrt 3}, \frac {1}{\sqrt 3}, \frac {1}{\sqrt 3}$
• Direction Cosines of AL are $-\frac {1}{\sqrt 3}, \frac {1}{\sqrt 3}, \frac {1}{\sqrt 3}$
• Direction Cosines of BM are $\frac {1}{\sqrt 3}, -\frac {1}{\sqrt 3}, \frac {1}{\sqrt 3}$
• Direction Cosines of CN are $\frac {1}{\sqrt 3}, \frac {1}{\sqrt 3}, -\frac {1}{\sqrt 3}$

Let us suppose l, m, n be the direction cosines of the line

So, line makes an angle $\Theta_1$ of  with OP.

Therefore, cos $\Theta_1$ =$l(\frac {1}{\sqrt 3}) + m(\frac {1}{\sqrt 3}) + n(\frac {1}{\sqrt 3})$

$cos \Theta_1 = \frac {l+m+n}{\sqrt 3}$ """(1)

Similarly, $cos \Theta_2 = \frac {-l+m+n}{\sqrt 3}$ """(2)

$cos \Theta_3 = \frac {l-m+n}{\sqrt 3}$ """(3)

$cos \Theta_4 = \frac {l+m-n}{\sqrt 3}$ """(4)

Squaring and Adding (1), (2), (3) and (4) we get-

= $cos^2\Theta_1 + cos^2\Theta_2 + cos^2\Theta_3 + cos^2\Theta_4 = \frac {1}{3}[(l+m+n)^2+(-l+m+n)^2+(l-m+n)^2+(l+m-n)^2]$

= $\frac {1}{3}[4l^2+4m^2+4n^2] = \frac {4}{3}(l^2+m^2+n^2) = \frac {4}{3}(1)$

So,$\ cos^2\Theta_1 + cos^2\Theta_2 + cos^2\Theta_3 + cos^2\Theta_4 = \frac {4}{3}$