Find a vector of magnitude 26 units normal to the plane 12x-3y+4z= 1.

Answers (1)

 The normal to this plane 12x-3y+4z= 1 is  \overrightarrow{\mathrm{n}}=12 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}

and Its magnitude is |\vec{n}|=\sqrt{(12)^{2}+(-3)^{2}+(4)^{2}}$=13

So Unit vector perpendicular to this plane is 

\hat{n}=\frac{12 \hat{i}-3 \hat{j}+4 \hat{k}}{13}=\frac{12}{13} \hat{1}-\frac{3}{13} \hat{j}+\frac{4}{13} \hat{k}

 

Now a vector normal to the plane with the magnitude 26 will be 

\begin{array}{l} =26 \hat{n} \\ \Rightarrow=26\left(\frac{12}{13} \hat{\imath}-\frac{3}{13} \hat{\jmath}+\frac{4}{13} \hat{k}\right) \\ \Rightarrow=24 \hat{\imath}-6 \hat{\jmath}+8 \hat{k} \end{array}

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