A) \[-24\hat{i}+32\hat{j}+30\hat{k}\]
B) \[24\hat{i}-32\hat{j}-30\hat{k}\]
C) \[12\hat{i}-16\hat{j}-15\hat{k}\]
D) \[-12\hat{i}+16\hat{j}-15\hat{k}\]
E) None of the above
Correct Answer: A
Solution :
Since,\[\overrightarrow{a}\]is collinear to vector\[\overrightarrow{b},\]therefore\[\overrightarrow{a}=m\overrightarrow{b}\]or some scalar m ie, \[\overrightarrow{a}=m\left( 6\hat{i}-8\hat{j}-\frac{15}{2}\hat{k} \right)\] \[\Rightarrow \] \[|\overrightarrow{a}|=|m|\sqrt{36+64+\frac{225}{4}}\] \[\Rightarrow \] \[50=\frac{25}{2}|m|\] \[\Rightarrow \] \[|m|=4\] \[\Rightarrow \] \[m=\pm 4\] Since,\[\overrightarrow{a}\]makes an acute angle with the positive direction of z-axis, so its z component must be positive and hence, W must be\[-4\]. \[\therefore \] \[\overrightarrow{a}=-4\left( 6\hat{i}-8\hat{j}-\frac{15}{2}\hat{k} \right)\] \[=-24\hat{i}+32\hat{j}+30\hat{k}\]
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