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Question

The perpendicular distance of the point $$P(1, 2, 3)$$ from the line $$\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}$$ is


A
7
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B
5
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C
0
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D
None
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Solution

The correct option is A $$7$$
We have,

$$\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}$$
Let point $$(1,2,3)$$ be $$P$$ and the point through which the line passes be $$Q(6,7,7).$$ 

Also, the line is parallel to the vector $$\vec{b}=3\hat{i}+2\hat{j}-2\hat{k}$$
Now,

$$\vec{PQ}=5\hat{i}+5\hat{j}+4\hat{k}$$

$$\therefore$$  $$\vec{b}\times \vec{PQ}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&2&-2\\5&5&4\end{vmatrix}$$

                   $$=\hat{i}(8+10)-\hat{j}(12+10)+\hat{k}(15-10)$$
                   $$=18\hat{i}-22\hat{j}+5\hat{k}$$
$$\Rightarrow$$  $$\left|\vec{b}\times \vec{PQ}\right|=\sqrt{(18)^2+(-22)^2+(5)^2}$$
                     
  $$=\sqrt{324+484+25}$$
                       $$=\sqrt{833}$$

$$\Rightarrow$$  $$d=\dfrac{\left|\vec{b}\times \vec{PQ}\right|}{\left|\vec{b}\right|}$$
         
 $$=\dfrac{|\sqrt{833|}}{|\sqrt{(3)^2+(2)^2+(2)^2|}}$$

          $$=\dfrac{\sqrt{833}}{\sqrt{17}}$$

          $$=\sqrt{49}$$

          $$=7$$

Mathematics

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