Question

If the coordinates of A and B be $\left( {1,2,3} \right)$ and $\left( {7,8,7} \right)$, then the projections of the line segment AB on the co-ordinate axis are
A. $\left( {6,6,4} \right)$
B. $\left( {4,6,4} \right)$
C. $\left( {3,3,2} \right)$
D. $\left( {2,3,2} \right)$

Answer 1

Hint: In order to solve this type of question, first we will find the position vector of AB. Then, by using the formula for finding the position vector, we will find the projection of AB on x, y and z-axis separately using the position vector obtained above. It will give us the projection of AB on the co-ordinate axis.

Formula used:
$\overrightarrow {AB} = $Position vector of B – Position vector of A

Complete step by step solution:
We are given,
$A\left( {1,2,3} \right)$ and $B\left( {7,8,7} \right)$
Finding the position vector of AB,
$\overrightarrow {AB} = $Position vector of B – Position vector of A
$\overrightarrow {AB} = \left( {7\widehat i + 8\widehat j + 7\widehat k} \right) - \left( {1\widehat i + 2\widehat j + 3\widehat k} \right)$
$\overrightarrow {AB} = 6\widehat i + 6\widehat j + 4\widehat k$
Let us find the projection of $\overrightarrow {AB} $ on x-axis,
\[\dfrac{{AB.\widehat i}}{{\left| i \right|}} = \dfrac{{6 \times 1}}{1} = 6\]
Projection of $\overrightarrow {AB} $ on y-axis,
\[\dfrac{{AB.\widehat j}}{{\left| j \right|}} = \dfrac{{6 \times 1}}{1} = 6\]
Projection of $\overrightarrow {AB} $ on z-axis,
\[\dfrac{{AB.\widehat k}}{{\left| k \right|}} = \dfrac{{4 \times 1}}{1} = 4\]
Thus, the projection on coordinate axis is $\left( {6,6,4} \right).$
$\therefore $ The correct option is A.

Note: The projection of (a, b, c) on x-axis is $\left( {a,0,0} \right)$, on y-axis is $\left( {0,b,0} \right)$ and on z-axis is $\left( {0,0,c} \right).$ Make sure to apply the formula correctly and solve it by substituting the values while finding the projection of AB on x, y and z-axis to get the desired correct answer.