Question

The \[ \bot \] distance of a corner of a unit cube on a diagonal not passing through is
A. \[\dfrac{{\sqrt 6 }}{3}\]
B. \[3\sqrt 3 \]
C. \[2\sqrt 3 \]
D. None of these

Answer 1

Hint: First we will first assume that the edges OA, OB, OC of a unit cube represents the unit vectors \[{\mathbf{i}}\], \[{\mathbf{j}}\], \[{\mathbf{k}}\] respectively and CM be the perpendicular from corner C on the diagonal OP to make a diagram. Then we will find the magnitude of the line segment OP using the formula \[\sqrt {{a^2} + {b^2} + {c^2}} \], where \[a\], \[b\], and \[c\] represents the coefficients of the unit vectors \[{\mathbf{i}}\], \[{\mathbf{j}}\], \[{\mathbf{k}}\] respectively to find the unit vector in the direction of \[\left| {\overrightarrow {{\text{OP}}} } \right|\] and then use the Pythagorean theorem \[{h^2} = {a^2} + {b^2}\], where \[h\] is the hypotenuse, \[a\] is the height and \[b\] is the base of the right-angled triangle to find the required value.

Complete step by step answer:

Let us assume that the edges OA, OB, OC of a unit cube represents the unit vectors \[{\mathbf{i}}\], \[{\mathbf{j}}\], \[{\mathbf{k}}\] respectively.

Let CM be the perpendicular from corner C on the diagonal OP.

Then, since it is a unit cube we have the line segment is \[\overrightarrow {{\text{OP}}} = {\mathbf{i}} + {\mathbf{j}} + {\mathbf{k}}\].

Finding the magnitude of the line segment OP using the formula \[\sqrt {{a^2} + {b^2} + {c^2}} \], where \[a\], \[b\], and \[c\] represents the coefficients of the unit vectors \[{\mathbf{i}}\], \[{\mathbf{j}}\], \[{\mathbf{k}}\] respectively, we get
\[
   \Rightarrow \left| {\overrightarrow {{\text{OP}}} } \right| = \sqrt {{1^2} + {1^2} + {1^2}} \\
   \Rightarrow \left| {\overrightarrow {{\text{OP}}} } \right| = \sqrt {1 + 1 + 1} \\
   \Rightarrow \left| {\overrightarrow {{\text{OP}}} } \right| = \sqrt 3 \\
 \]

Dividing the line segment by OP by its magnitude the find the unit vector in the direction of\[\left| {\overrightarrow {{\text{OP}}} } \right|\], we get

\[ \Rightarrow \dfrac{{{\mathbf{i}} + {\mathbf{j}} + {\mathbf{k}}}}{{\sqrt 3 }}\]

We know that when the diagonals intersect each other in a square, that is, the line segment OM is the projection of side of the square OC on OP from the given diagram, then we have
\[{\text{OM}} = \overrightarrow {{\text{OC}}} \cdot \left( {{\text{unit vector along }}\overrightarrow {{\text{OP}}} } \right)\].

Substituting the value of OC and the above unit vector along \[\left| {\overrightarrow {{\text{OP}}} } \right|\] in the above equation, we get
\[
   \Rightarrow {\text{OM}} = {\mathbf{j}} \cdot \left( {\dfrac{{{\mathbf{i}} + {\mathbf{j}} + {\mathbf{k}}}}{{\sqrt 3 }}} \right) \\
   \Rightarrow {\text{OM}} = 0 + \dfrac{1}{{\sqrt 3 }} + 0 \\
   \Rightarrow {\text{OM}} = \dfrac{1}{{\sqrt 3 }} \\
 \]

We will use the Pythagorean theorem \[{h^2} = {a^2} + {b^2}\], where \[h\] is the hypotenuse, \[a\] is the height and \[b\] is the base of the right-angled triangle.

Applying the Pythagorean theorem in the triangle OCM in the above figure, we get

\[
   \Rightarrow {\text{O}}{{\text{C}}^2} = {\text{C}}{{\text{M}}^2} + {\text{O}}{{\text{M}}^2} \\
   \Rightarrow {1^2} = {\text{C}}{{\text{M}}^2} + {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2} \\
   \Rightarrow 1 = {\text{C}}{{\text{M}}^2} + \dfrac{1}{3} \\
 \]

Subtracting the above equation by \[\dfrac{1}{3}\] on each of the sides, we get

\[
   \Rightarrow 1 - \dfrac{1}{3} = {\text{C}}{{\text{M}}^2} \\
   \Rightarrow \dfrac{{3 - 1}}{3} = {\text{C}}{{\text{M}}^2} \\
   \Rightarrow {\text{C}}{{\text{M}}^2} = \dfrac{2}{3} \\
 \]

Taking the square root on both sides in the above equation, we get

\[ \Rightarrow {\text{CM}} = \pm \sqrt {\dfrac{2}{3}} \]

Since the side can never be negative, the negative value of \[{\text{CM}}\] is discarded.

So, the perpendicular distance is \[\sqrt {\dfrac{2}{3}} \].

Since options A, B and C do not match with the final answer, so option D is correct.

Note: In solving these types of questions, students should make a diagram for better understanding and label the vertices properly to avoid confusion. One should know that the magnitude of a vector is the length of a line segment and the vector, which has a magnitude 1 is known as the unit vector. Do not use the line segment OP instead of finding the unit vector of OP.