Question

If a vector $\vec a$ of magnitude 50 is collinear with vector $\vec b = 6\hat i - 8\hat j - \dfrac{{15}}{2}\hat k$ and makes an acute angle with positive z- axis then:
A) $\vec a = 4\vec b$
B) $\vec a = - 4\vec b$
C) $\vec b = 4\vec a$
D) None

Answer 1

We can find the unit vector along the direction of vector b by dividing it with its magnitude. Then we can find the unit vector of vector $\vec a$ by changing the sign of the unit vector of b by making the z component positive using the condition of the angle. Then we can multiply the unit vector with the magnitude of a to get the required vector. Then we can find the required relation.

Complete step by step solution:
We have the vector $\vec b = 6\hat i - 8\hat j - \dfrac{{15}}{2}\hat k$.
We can find its magnitude. We know that magnitude of a vector $P = a\hat i + b\hat j + c\hat k$ is given by the equation, $\left| P \right| = \sqrt {{a^2} + {b^2} + {c^2}}$
So, the magnitude of vector b is given by,
$\Rightarrow \left| {\vec b} \right| = \sqrt {{6^2} + {{\left( { - 8} \right)}^2} + {{\left( { - \dfrac{{15}}{2}} \right)}^2}}$
On simplification, we get,
$\Rightarrow \left| {\vec b} \right| = \sqrt {36 + 64 + \dfrac{{225}}{4}}$
We can add the integer terms.
$\Rightarrow \left| {\vec b} \right| = \sqrt {100 + \dfrac{{225}}{4}}$
Now we can take the LCM.
$\Rightarrow \left| {\vec b} \right| = \sqrt {\dfrac{{400 + 225}}{4}}$
On adding the numerator, we get,
$\Rightarrow \left| {\vec b} \right| = \sqrt {\dfrac{{625}}{4}}$
On taking the square root, we get,
$\Rightarrow \left| {\vec b} \right| = \dfrac{{25}}{2}$
Now we can find the unit vector along the direction of b.
$\hat b = \dfrac{{\vec b}}{{\left| {\vec b} \right|}}$
On substituting the values, we get,
$\Rightarrow \hat b = \dfrac{{6\hat i - 8\hat j - \dfrac{{15}}{2}\hat k}}{{\dfrac{{25}}{2}}}$
On taking the denominator to the numerator taking the reciprocal, we get,
$\Rightarrow \hat b = \dfrac{2}{{25}}\left( {6\hat i - 8\hat j - \dfrac{{15}}{2}\hat k} \right)$
Now as the vectors a and b are collinear, So, a can be in the direction of $\hat b$ or $- \hat b$ .
It is given that the vector makes an acute angle with the positive z axis. So, the 3rd direction cosine will be positive. It is negative for $\hat b$ . So, a is in the direction of $- \hat b$
So, we can write the unit vector of a as,
$\Rightarrow \hat a = - \hat b$
On substituting, we get,
$\Rightarrow \hat a = - \dfrac{2}{{25}}\left( {6\hat i - 8\hat j - \dfrac{{15}}{2}\hat k} \right)$
It is given that the magnitude of vector a is 50. So, the vector will be equal to 50 times the unit vector.
$\Rightarrow \vec a = 50 \times \hat a$
$\Rightarrow \vec a = 50 \times - \dfrac{2}{{25}}\left( {6\hat i - 8\hat j - \dfrac{{15}}{2}\hat k} \right)$
On simplification, we get
$\Rightarrow \vec a = - \dfrac{{50 \times 2}}{{25}}\left( {6\hat i - 8\hat j - \dfrac{{15}}{2}\hat k} \right)$
On multiplication and division of the term outside the bracket, we get,
$\Rightarrow \vec a = - 4\left( {6\hat i - 8\hat j - \dfrac{{15}}{2}\hat k} \right)$
The term inside the bracket is vector b. so we can write,
$\Rightarrow \vec a = - 4\vec b$
Thus, the required relation is $\vec a = - 4\vec b$ .

So, the correct answer is option B.

Note:
We need not to find the exact values of the vectors and unit vectors in the intermediate steps. It is time consuming and it is again difficult to convert it in terms of vector b. We must note that collinear vectors are vectors that lie on a line. So, the vector can be either parallel or antiparallel. The direction cosines of a vector are the cosines of the angle it makes with the positive direction of the axes. We took the case where the vector component in the direction of z axis is positive because cosines of acute angles are positive always as they lie in the $1^{\text{st}}$ quadrant.