Question

Find the angle between Diagonals of a Cube.

Class 12th

Vector and 3D ​

Answer 1

Solution

Let,O be the vertex of a cube .....

now...

so the three edges through the point O are the axes (X,Y & Z axis),

Now....

the four diagonals are OP,AA',BB' and CC'.

now..let the edge of the cube is= a unit.

therefore the coordinates of P,A and A are ...

(a,a,a);(a,0,0) & (0,a,a)

therefore the direction cosine ratios of OP are

a,a,a

therefore the direction cosines of OP are

$= \frac{a}{a \sqrt{3} } , \frac{a}{a \sqrt{3} } , \frac{a}{a \sqrt{3} } \\= \frac{1}{ \sqrt{3} } , \frac{1}{ \sqrt{3} } , \frac{1}{ \sqrt{3} }$

similarly.....

direction cosine of AA'

$= \frac{-1}{ \sqrt{3} } , \frac{1}{ \sqrt{3} } , \frac{1}{ \sqrt{3} }$

let , "a" be the angle between OP and AA'

therefore...

$\cos {}^{ - 1} (a) = ( \frac{1}{ \sqrt{3} } )( \frac{ - 1}{3} ) + ( \frac{1}{ \sqrt{3} } )( \frac{1}{ \sqrt{3} } ) + ( \frac{1}{ \sqrt{3} } )( \frac{1}{ \sqrt{3} } ) \\ = > cos (a) = \frac{1}{3} \\ = > a = \cos {}^{ - 1} ( \frac{1}{3} )$

Answer 2

Let,O be the vertex of a cube

....now.

..so the three edges through the point O are the axes (X,Y & Z axis),

Now.

...the four diagonals are OP,AA',BB' and CC'.

now.

.let the edge of the cube is= a unit.therefore the coordinates of P,A and A are ...(a,a,a);(a,0,0) & (0,a,a)therefore the direction cosine ratios of OP are

next........ refers to the attachment