Question

the vector parallel to 4icap_3jcap+5kcap and wh9se length is the artematic mean of lengths of two vectors 2 icap - 4jcap=4kcapand icap+root6jcap+3kcap​

Answer 1

Correct Question:-

Find the vector parallel to $\bf{4\ \hat i-3\ \hat j+5\ \hat k}$ and whose length is the arithmetic mean of lengths of two vectors $\bf{2\ \hat i-4\ \hat j+4\ \hat k}$ and $\bf{\hat i+\sqrt6\ \hat j+3\ \hat k}.$

Solution:-

Length of the vector $\bf{2\ \hat i-4\ \hat j+4\ \hat k},$

$\sf{=\sqrt{2^2+(-4)^2+4^2}}$

$\sf{=\sqrt{4+16+16}}$

$\sf{=\sqrt{36}}$

$\sf{=6}$

Length of the vector $\bf{\hat i+\sqrt6\ \hat j+3\ \hat k},$

$\sf{=\sqrt{1^2+(\sqrt6)^2+3^2}}$

$\sf{=\sqrt{1+6+9}}$

$\sf{=\sqrt{16}}$

$\sf{=4}$

The arithmetic mean of these lengths $\sf{=\dfrac{6+4}{2}=5.}$

This is the length of the vector to be found.

Length of the vector $\bf{4\ \hat i-3\ \hat j+5\ \hat k},$

$\sf{=\sqrt{4^2+(-3)^2+5^2}}$

$\sf{=\sqrt{16+9+25}}$

$\sf{=\sqrt{50}}$

$\sf{=5\sqrt2}$

Then, the vector to be found is given by,

$\longrightarrow\vec{\sf{a}}=\sf{\dfrac{5}{5\sqrt2}}\,\left(\bf{4\ \hat i-3\ \hat j+5\ \hat k}\right)$

$\longrightarrow\vec{\sf{a}}=\sf{\dfrac{1}{\sqrt2}}\,\left(\bf{4\ \hat i-3\ \hat j+5\ \hat k}\right)$

$\longrightarrow\underline{\underline{\vec{\sf{a}}=\bf{2\sqrt2\ \hat i-\dfrac{3}{2}\sqrt2\ \hat j+\dfrac{5}{2}\sqrt2\ \hat k}}}$