Question

The projection of the join of the two points $\left( 1,4,5 \right)$ , $\left( 6,7,2 \right)$ on the line whose d.s’s are
$\left( 4,5,6 \right)$
(A) $\dfrac{17}{\sqrt{77}}$
(B) $\dfrac{7}{6}$
(C) 21
(D) $\dfrac{7}{9}$

Answer 1

Hint: Convert the cartesian form of the point A which has coordinate $\left( 1,4,5 \right)$ and point B which has coordinate $\left( 6,7,2 \right)$ into vector form. Now, we have $\overrightarrow{A}=1\widehat{i}+4\widehat{j}+5\widehat{k}$ and $\overrightarrow{B}=6\widehat{i}+7\widehat{j}+2\widehat{k}$ . Now, get the value of $\overrightarrow{AB}$ by putting the value of $\overrightarrow{A}$ and $\overrightarrow{B}$ in the equation, $\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}$ . We know the formula for the projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ , $\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{b} \right|}$ . Now, replace $\overrightarrow{a}$ by $\overrightarrow{AB}$ and $\overrightarrow{b}$ by $\overrightarrow{C}=4\widehat{i}+5\widehat{j}+6\widehat{k}$ in the formula for the projection and then, solve it further.

Complete step-by-step answer:
According to the question, it is given that we have the coordinate of two points and the coordinates are $\left( 1,4,5 \right)$ and $\left( 6,7,2 \right)$ . We also have a vector whose direction ratios are $\left( 4,5,6 \right)$ .
Let us assume we have two points A and B whose coordinates are $\left( 1,4,5 \right)$ and $\left( 6,7,2 \right)$ respectively.
Now, converting the coordinates of the points A and B into vector form, we get
$\overrightarrow{A}=1\widehat{i}+4\widehat{j}+5\widehat{k}$ …………………..(1)
$\overrightarrow{B}=6\widehat{i}+7\widehat{j}+2\widehat{k}$ ……………………(2)
It is also given that we have a vector whose direction ratio is $\left( 4,5,6 \right)$ . Let us assume a vector $\overrightarrow{C}$ whose direction ratio is $\left( 4,5,6 \right)$ .
Now, using the direction ratio $\left( 4,5,6 \right)$ , we get
$\overrightarrow{C}=4\widehat{i}+5\widehat{j}+6\widehat{k}$ ………………………(3)

Now, from the figure, we have, $\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}$ ……………………(4)
From equation (1) and equation (2), we have vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ .
Putting the value of $\overrightarrow{A}$ and $\overrightarrow{B}$ in equation (4), we get

& \overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A} \\\ & \Rightarrow \overrightarrow{AB}=\left( 6\widehat{i}+7\widehat{j}+2\widehat{k} \right)-\left( 1\widehat{i}+4\widehat{j}+5\widehat{k} \right) \\\ & \Rightarrow \overrightarrow{AB}=\left( 6-1 \right)\widehat{i}+\left( 7-4 \right)\widehat{j}+\left( 2-5 \right)\widehat{k} \\\ \end{aligned}$$ $$\Rightarrow \overrightarrow{AB}=5\widehat{i}+3\widehat{j}+\left( -3 \right)\widehat{k}$$ ……………………….(5) Now, we have to find the projection of $$\overrightarrow{AB}$$ on $$\overrightarrow{C}$$ . We know the formula for the projection of $$\overrightarrow{a}$$ on $$\overrightarrow{b}$$ , $$\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{b} \right|}$$ ……………….(6) Now, using this formula to find the projection of $$\overrightarrow{AB}$$ on $$\overrightarrow{C}$$ .  From equation (3) and equation (5), we have the value of $$\overrightarrow{AB}$$ and $$\overrightarrow{C}$$ . The dot product of two vector is obtained by summation of the multiplication of the z-component of one vector with the z-component of the other vector, multiplication of the y-component of one vector with the y-component of the other vector, and the multiplication of the z-component of one vector with the z-component of the other vector. Now, replacing $$\overrightarrow{a}$$ by $$\overrightarrow{AB}$$ and $$\overrightarrow{b}$$ by $$\overrightarrow{C}$$ in equation (4), we get The projection of $$\overrightarrow{AB}$$ on $$\overrightarrow{C}$$ = $$\dfrac{\overrightarrow{AB}.\overrightarrow{C}}{\left| \overrightarrow{AB} \right|}=\dfrac{\left( 5\widehat{i}+3\widehat{j}-3\widehat{k} \right).\left( 4\widehat{i}+5\widehat{j}+6\widehat{k} \right)}{\left| \left( 4\widehat{i}+5\widehat{j}+6\widehat{k} \right) \right|}=\dfrac{20+15-18}{\sqrt{16+25+36}}=\dfrac{17}{\sqrt{77}}$$ . Therefore, the projection of $$\overrightarrow{AB}$$ on $$\overrightarrow{C}$$ is $$\dfrac{17}{\sqrt{77}}$$ . Hence, the correct option is (A). Note: For this question, one might think to solve it in cartesian form only. If we try to solve it in cartesian form, then it will become very complex to solve because it is very difficult to imagine and work in a three coordinate axis. To reduce the complexity, we must convert the cartesian coordinates into vector form and in vector form it will be easy to operate and solve. Also, in this question the hidden information is the direction ratio of the vector $$\overrightarrow{C}$$ which is $$\left( 4,5,6 \right)$$ .