Question

if a vector a of magnitude 50 is collinear with b=6i-8j-7.5k and makes an acute angle with positive z axis then a=?

Answer 1

b
b=+or-50/12.5(6i-8j-7.5k)
Z=pk
b. K is greater than 0
Therefore b=-4(6i-8j-7.5k)
b=-4a

Answer 2

Answer is $\overrightarrow{a} = -4\overrightarrow{b}$

Step-by-step explanation:

The given vector b= $6\overrightarrow{i} - 8\overrightarrow{j} - \frac {15}{2}\overrightarrow{k}$

the unit vector along b will be

= $\pm \frac {6\overrightarrow{i} - 8\overrightarrow{j} - \frac {15}{2}\overrightarrow{k}}{\sqrt {6^2+8^2+\frac {15^2}{2^2}}}$

= $\pm \frac {6\overrightarrow{i} - 8\overrightarrow{j} - \frac {15}{2}\overrightarrow{k}}{\sqrt {36+64+\frac {225}{4}}}$

= $\pm \frac {6\overrightarrow{i} - 8\overrightarrow{j} - \frac {15}{2}\overrightarrow{k}}{\sqrt {\frac{36\times 4+64\times 4+225}{4}}}$

= $\pm \frac {6\overrightarrow{i} - 8\overrightarrow{j} - \frac {15}{2}\overrightarrow{k}}{\sqrt {\frac{144+256+225}{4}}}$

= $\pm \frac {6\overrightarrow{i} - 8\overrightarrow{j} - \frac {15}{2}\overrightarrow{k}}{\sqrt {\frac{625}{4}}}$

= $\pm \frac {2}{25}\times (6\overrightarrow{i} - 8\overrightarrow{j} - \frac {15}{2}\overrightarrow{k})$

Now the vector of magnitude 50 along b

a = $\pm 50 \times \frac {2}{25}\times (6\overrightarrow{i} - 8\overrightarrow{j} - \frac {15}{2}\overrightarrow{k})$

a = $\pm 4 \times (6\overrightarrow{i} - 8\overrightarrow{j} - \frac {15}{2}\overrightarrow{k})$

now as the vector makes acute angle with z−axis,  

we must have a.k > 0 neglecting the positive value as a.k > 0 only for negative value

a = $\pm 4 \times (6\overrightarrow{i} - 8\overrightarrow{j} - \frac {15}{2}\overrightarrow{k})$

$\overrightarrow{a} = -4\overrightarrow{b}$