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  • Find a vector of magnitude 6, which is perpendicular to both the vectors 2i minus j plus 2k and 4i minus j plus 3k

Find a vector of magnitude 6, which is perpendicular to both the vectors 2 \hat{i}-\hat{j}+2 \hat{k}  and 4\hat{i}-\hat{j}+3 \hat{k}

Answers (1)

Let the vectors be \\ \vec{a} and \\ \vec{b} , such that

\\ \vec{a}=2 \hat{\imath}-\hat{\jmath}+2 \hat{k} \\ \vec{b}=4 \hat{\imath}-\hat{\jmath}+3 \hat{k}

We need to find a vector perpendicular to both the vectors  \\ \vec{a} and \\ \vec{b}

 

Any vector perpendicular to both \\ \vec{a} and \\ \vec{b}  can be given as,

\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -1 & 2 \\ 4 & -1 & 3 \end{array}\right| \\ \begin{aligned} \Rightarrow \vec{a} \times \vec{b}=\hat{i}((-1)(3)-(2)(-1))-\hat{j}((2)(3)-(2)(4)) \\ +\hat{k}((2)(-1)-(-1)(4)) \end{aligned} \\ \Rightarrow \vec{a} \times \overrightarrow{\mathrm{b}}=\hat{\imath}(-3+2)-\hat{\jmath}(6-8)+\hat{\mathrm{k}}(-2+4) \\ \Rightarrow \vec{a} \times \overrightarrow{\mathrm{b}}=-\hat{\imath}+2 \hat{\jmath}+2 \hat{\mathrm{k}}

Let

\\ \overrightarrow{\mathrm{r}}=-\hat{\mathrm{l}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}
 

As we know, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.

So,

 

\begin{align*} &\text{A vector of magnitude 6 in the direction of } \mathbf{r} \text{ is given by,}\\ &\text{vector} = 6 \times \frac{\mathbf{r}}{|\mathbf{r}|} \\ &\Rightarrow \text{vector} = 6 \times \frac{-\hat{\imath}+2\hat{\jmath}+2\hat{k}}{|- \hat{\imath}+2\hat{\jmath}+2\hat{k}|} \\ &\text{Here, } |- \hat{\imath}+2\hat{\jmath}+2\hat{k}| = \sqrt{(-1)^{2}+(2)^{2}+(2)^{2}} \end{align*}

\\ \Rightarrow \text { vector }=6 \times \frac{-\hat{1}+2 \hat{\jmath}+2 \hat{\mathrm{k}}}{\sqrt{(-1)^{2}+(2)^{2}+(2)^{2}}} \\ \Rightarrow \text { vector }=6 \times \frac{(-\hat{\imath}+2 \hat{\jmath}+2 \hat{\mathrm{k}})}{\sqrt{1+4+4}} \\ \Rightarrow \text { vector }=6 \times \frac{-\hat{\mathrm{i}}+2 \hat{\jmath}+2 \hat{\mathrm{k}}}{\sqrt{9}}

\\ \begin{aligned} &\Rightarrow \text { vector }=6 \times \frac{-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3}\\ &\Rightarrow \text { vector }=2 \times(-\hat{1}+2 \hat{\jmath}+2 \hat{\mathrm{k}})\\ &\Rightarrow \text { vector }=-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k}\\ &\text { Thus, required vector is }-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k} \end{aligned}

 

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