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  • The vector in the direction of the vector i -2j +2k that has magnitude 9 is A. i -2j +2k

The vector in the direction of the vector \hat{i}-2 \hat{j}+2 \hat{k}  that has magnitude 9 is


\\ A. \hat{i}-2 \hat{j}+2 \hat{k}\\ B.\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}\\ C.3(\hat{i}-2 \hat{j}+2 \hat{k})\\ D. 9(\hat{i}-2 \hat{j}+2 \hat{k})\\

 

Answers (1)

C)

Given is the vector   \hat{i}-2 \hat{j}+2 \hat{k}

Let this vector be \vec{a} , such that

\vec{a}=\hat{i}-2 \hat{j}+2 \hat{k}

Let us first find the unit vector in the direction of this vector \vec{a} .

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.

Unit vector in the direction of the vector \vec{a}  is given as,

\\ \begin{aligned} &\hat{a}=\frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}\\ &\text { As, we have } \overrightarrow{\mathrm{a}}=\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}\\ &\text { Then, }\\ &|\overrightarrow{\mathrm{a}}|=|\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}|\\ &\Rightarrow|\vec{a}|=\sqrt{1^{2}+(-2)^{2}+2^{2}}\\ &[\because \text { if }|\vec{p}|=|x \hat{\imath}+y \hat{\jmath}+z \hat{k}| \end{aligned}

\\ \Rightarrow|\vec{p}|=\sqrt{x^{2}+y^{2}+z^{2}} \\ \Rightarrow|\vec{a}|=\sqrt{1+4+4} \\ \Rightarrow|\vec{a}|=\sqrt{9} \\ \Rightarrow|\vec{a}|=3

Therefore,

\\ \hat{\mathrm{a}}=\frac{\hat{1}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3} \\ {\left[\because \overrightarrow{\mathrm{a}}=\hat{1}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}_{\text {and }}|\overrightarrow{\mathrm{a}}|=3 \right].}

We have found unit vector in the direction of the vector \hat{i}-2 \hat{j}+2 \hat{k} , but we need to find the unit vector in the direction of \hat{i}-2 \hat{j}+2 \hat{k}  but also with the magnitude 9.

We have the formula:

Vector in the direction of  \vec{a} with a magnitude of 9=9 \times \frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}

\\ \begin{aligned} &=9 \times \widehat{a}\\ &\text { And } \hat{\mathrm{a}}=\frac{\hat{1}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3} \text { as just found. }\\ &\text { So, }\\ &\Rightarrow \text { Vector in the direction of } \overrightarrow{\mathrm{a}} \text { with a magnitude of } 9=9 \times \frac{\hat{1}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3} \end{aligned}

\\ \begin{aligned} &=3(\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}})\\ &\text { Thus, vector in the direction of vector }\\ &\hat{\imath}-2 \hat{\jmath}+2 \hat{k}_{\text {and }} \text { has magnitude } 9 \text { is } 3(\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}) \end{aligned}

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