Question

If a=2i+2j-kand b=i+j+k . Find a vector c which is parallel to a but has magnitude three times that of b

Answer 1

Vector $\vec c$ which is parallel to  $\vec a$ but has magnitude three times that of  $\vec b$ is

$\boxed{\sqrt{3}(2\hat i+2\hat j-\hat k)}$

Explanation:

Given

$\vec a=2\hat i+2\hat j-\hat k$

$\vec b=\hat i+\jat j+\hat k$

If a vector $\vec c$ is parallel to $\vec a$

Then,

$\vec c=\lambda \vec a$

$\implies \vec c=\lambda(2\hat i+2\hat j-\hat k)$

Magnitude of $\vec c$

$|\vec c|=\sqrt{\lambda^2(2^2+2^2+(-1)^2)}$

$\implies |\vec c|=\sqrt{\lambda^2(9}$

$\implies |\vec c|=3\lambda$

Magnitude of $\vec b$

$|\vec b|=\sqrt{1^2+1^2+1^2}$

$\implies |\vec b|=\sqrt{3}$

Given that

$|\vec c|=3|\vec b|$

$\implies 3\lambda=3\sqrt{3}$

$\implies \lambda=\sqrt{3}$

Therefore,  $\vec c$ is

$\vec c=\sqrt{3}(2\hat i+2\hat j-\hat k)$

Hope this answer is helpful.

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