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  • Find a unit vector in the direction of PQ , where P and Q have co-ordinates (5, 0, 8) and (3, 3, 2), respectively.

Find a unit vector in the direction of  \bar{PQ} , where P and Q have co-ordinates (5, 0, 8) and (3, 3, 2), respectively.

Answers (1)

We have,

Coordinates of P is (5, 0, 8).

Coordinates of Q is (3, 3, 2).

So,

Position vector of P is given by,

\\ \begin{aligned} &\overrightarrow{\mathrm{OP}}=5 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OP}}=5 \hat{\mathrm{i}}+8 \hat{\mathrm{k}}\\ &\text { Position vector of } Q \text { is given by, }\\ &\overrightarrow{\mathrm{OQ}}=3 \hat{\imath}+3 \hat{\jmath}+2 \hat{\mathrm{k}} \end{aligned}

To find unit vector in the direction of PQ, we need to find position vector of PQ.

Position vector of PQ is given by,

\\ \overrightarrow{\mathrm{PQ}}=\text { Position vector of } \mathrm{Q}-\text { Position vector of } \mathrm{P} \\ \Rightarrow \overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}} \\ \Rightarrow \overrightarrow{\mathrm{PQ}}=(3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})-(5 \hat{\mathrm{l}}+8 \hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{PQ}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}-8 \hat{\mathrm{k}}

\Rightarrow \overrightarrow{\mathrm{PQ}}=-2 \hat{\imath}+3 \hat{\jmath}-6 \hat{\mathrm{k}}

We know that a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.

To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.

For finding unit vector, we have the formula:

\\ \widehat{\mathrm{PQ}}=\frac{\overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{PQ}}|} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\imath}+3 \hat{\jmath}-6 \hat{\mathrm{k}}}{|-2 \hat{\imath}+3 \hat{\jmath}-6 \hat{\mathrm{k}}|} \\ \text { Here, }|-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}|=\sqrt{(-2)^{2}+3^{2}+(-6)^{2}} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{\sqrt{(-2)^{2}+3^{2}+(-6)^{2}}}

\\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\imath}+3 \hat{\jmath}-6 \hat{\mathrm{k}}}{\sqrt{4+9+36}} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\mathrm{l}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{\sqrt{49}} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\mathrm{l}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}

Thus, unit vector in the direction of PQ is \frac{-2 \hat{\mathrm{l}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}

Posted by

infoexpert22

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