The angle between any two diagonals of a cube | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the side length of the cube be $a$. We consider a cube with vertices at $(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a),$ and $

Vedclass · Accepted Answer

$\cos ^{-1}\left(\frac{1}{3}\right)$

Vedclass · Answer

(A) Let the side length of the cube be $a$. We consider a cube with vertices at $(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a),$ and $(a,a,a)$.Two diagonals of the cube can be represented by vectors $\vec{d_1}$ and $\vec{d_2}$.Let $\vec{d_1}$ be the vector from $(0,0,0)$ to $(a,a,a)$,so $\vec{d_1} = a\hat{i} + a\hat{j} + a\hat{k}$.Let $\vec{d_2}$ be the vector from $(a,0,0)$ to $(0,a,a)$,so $\vec{d_2} = (0-a)\hat{i} + (a-0)\hat{j} + (a-0)\hat{k} = -a\hat{i} + a\hat{j} + a\hat{k}$.The angle $	heta$ between these two vectors is given by $\cos 	heta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|}$.Calculating the dot product: $\vec{d_1} \cdot \vec{d_2} = (a)(-a) + (a)(a) + (a)(a) = -a^2 + a^2 + a^2 = a^2$.Calculating the magnitudes: $|\vec{d_1}| = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3}$ and $|\vec{d_2}| = \sqrt{(-a)^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3}$.Thus,$\cos 	heta = \frac{a^2}{(a\sqrt{3})(a\sqrt{3})} = \frac{a^2}{3a^2} = \frac{1}{3}$.Therefore,$	heta = \cos ^{-1}\left(\frac{1}{3}ight)$.

The angle between any two diagonals of a cube is

Similar Questions

If the position vectors of the vertices of $\Delta ABC$ are $2\hat{i} + 4\hat{j} - \hat{k}$,$4\hat{i} + 5\hat{j} + \hat{k}$,and $3\hat{i} + 6\hat{j} - 3\hat{k}$,then which of the following angles is a right angle?

If the non-zero vectors $a$ and $b$ are perpendicular to each other,then the solution of the equation $r \times a = b$ is given by

If $A(3,4,5), B(4,6,3), C(-1,2,4)$ and $D(1,0,5)$ are such that the angle between the lines $DC$ and $AB$ is $\theta$,then $\cos \theta$ is equal to

In quadrilateral $ABCD$,$AB=a$,$BC=b$,$AD=b-a$. If $M$ is the midpoint of $BC$ and $N$ is a point on $DM$ such that $DN=\left(\frac{4}{5}\right) DM$,then $5 AN=$

The value of $\frac{(\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2}{2|\vec{a}|^2|\vec{b}|^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module