If {a} is a vector of magnitude 50 , collinear with the vector {b}=6 {i}-8 {j}-15/2 {k} and makes an acute angle with the positive direction of Z - axis, then {a} is equal to

Correct Answer is : −24i^+32j^+30k^-24 {i}+32 {j}+30 {k}−24i^+32j^+30k^

Correct Answer is : −24i^+32j^+30k^-24 {i}+32 {j}+30 {k}−24i^+32j^+30k^

Test Index

BITSAT 2021 Paper

Section: Mathematics

Question : 145 of 150

Marks: +1, -0

\vec{a}

\vec{b}=6 \hat{i}-8 \hat{j}-\frac{15}{2} \hat{k}

\vec{a}

is equal to

Solution:

Since

\vec{a}=m \vec{b}

for some scalar

m

i.e.,

\vec{a}=m\left(6 \hat{i}-8 \hat{j}-\frac{15}{2} \hat{k}\right)

⇒

\left|a\right|=\left|m\right|\sqrt{36+64+\frac{225}{4}}

⇒

50=\frac{25}{2}\left|m\right|

⇒

\left|m\right|=4

⇒

m=\pm 4

Since, a makes an acute angle with the positive direction of Z - axis, so its

z

component must be positive and hence,

m

must be

-4

∴

a=-4\left(6 \hat{i}+8 \hat{j}-\frac{15}{2} \hat{k}\right)

=-24 \hat{i}+32 \hat{j}+30 \hat{k}

Go to Question:

Prev Question

Next Question