MCQ
The angle between two diagonals of a cube will be
- A${\sin ^{ - 1}}1/3$
- ✓${\cos ^{ - 1}}1/3$
- CVariable
- DNone of these
✓
Answer
Correct option: B.
${\cos ^{ - 1}}1/3$
b
(b) Let the cube be of side $'a'$
(b) Let the cube be of side $'a'$
$O\,(0,\,\,0,\,\,0),\,\,D\,(a,\,\,a,\,\,a),\,\,B\,(0,\,\,a,\,\,0),\,\,G\,(a,\,\,0,\,\,a)$
Then equation of $OD$ and $BG$ are $\frac{x}{a} = \frac{y}{a} = \frac{z}{a}$ and
$\frac{x}{a} = \frac{{y - a}}{{ - a}} = \frac{z}{a}$ respectively.
Hence, angle between $OD$ and $BG$ is
${\cos ^{ - 1}}\left( {\frac{{{a^2} - {a^2} + {a^2}}}{{\sqrt {3{a^2}} .\,\sqrt {3{a^2}} }}} \right) = {\cos ^{ - 1}}\,\left( {\frac{1}{3}} \right)$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The value of ${d \over {dx}}[|x - 1| + |x - 5|]$ at $x = 3$ is
→$\int_{}^{} {x{{\tan }^{ - 1}}} xdx = $
→If $y = {\tan ^{ - 1}}\left( {{x \over {1 + \sqrt {1 - {x^2}} }}} \right)$, then ${{dy} \over {dx}} = $
→Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non $-$ zero vectors such that $\vec{b}$ and $\vec{c}$ are non $-$ collinear if $\vec{a}+5 \vec{b}$ is collinear with $\overrightarrow{ c }, \overrightarrow{ b }+6 \overrightarrow{ c }$ is collinear with $\overrightarrow{ a }$ and $\overrightarrow{ a }+\alpha \overrightarrow{ b }+\beta \overrightarrow{ c }=\overrightarrow{0}$, then $\alpha+\beta$ is equal to
→The locus of the centre of a circle which cuts orthogonally the circle ${x^2} + {y^2} - 20x + 4 = 0$ and which touches $x = 2$ is
→Two real numbers $\alpha$ and $\beta $ are such that $\alpha + \beta = 3$ and $\left| {\alpha - \beta } \right| = 4$, then $\alpha$ and $\beta $ are the roots of the quadratic equation
→Which pair $(s)$ of function $(s)$ is/are equal ?
→where $\{x\}$ and $[x]$ denotes the fractional part $\&$ integral part functions.
Given vertices $A(1,\,1),B(4,\, - 2)$ and $C(5,\,5)$ of a triangle, then the equation of the perpendicular dropped from $C$ to the interior bisector of the angle $A$ is
→Let $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha\end{array}\right]$ and $|2 A|^3=2^{21}$ where $\alpha, \beta \in Z,$ Then a value of $\alpha$ is
→If $A = \{ 1,\,2,\,3,\,4\} $; $B = \{ a,\,b\} $ and $f$ is a mapping such that $f:A \to B$, then $A \times B$ is
→