Question

if 0.2i+0.3j+zk is a unit vector then zsqaureis

Answer 1

it is a unit vector so it's magnitude is one
so
√[(0.2)^2 + (0.3)^2 + z^2] = 1
0.04 + 0.09 + z^2 = 1
z^2= 1- 0.13
z^2= 0.87

Answer 2

Answer:

The value of $z^2$ is $0.87.$

Explanation:

Step 1 : Let us assume vector that $P = 0.2i + 0.3j+zk$.

If vector P be a unit vector that magnitude of vector P be 1 .

Step 2 : Magnitude of any vector of form $a i + bj +ck$  is given by $\sqrt{a^{2}+b^{2} +c^{2} }$ .

$|P |= 1\\ \sqrt{0.2^{2} + 0.3^{2} + z^{2} } = 1\\ \sqrt{0.04 + 0.09 + z ^{2} } = 1$

on squaring both sides,

$0.04+0.09+z^{2} = 1\\0.13 +z^{2} = 1\\z^{2} = 1 - 0.13\\z^{2} = 0.87$

Thus,the final answer is 0.87.

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