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Answer 1 · 2017-09-11T10:29:21

$∣ ∣ ∣ \to A ∣ ∣ ∣ = \sqrt{106}$ $|vec("A")| = sqrt(106)$

The vector $A$ $"A"$ has components $a_{x} = - 5$ $a_(x) = - 5$ $m$ $"m"$ and $a_{y} = 9$ $a_(y) = 9$ $m$ $"m"$ .

This can be written as:

$\Rightarrow \to A = - 5 i + 9 j$ $Rightarrow vec("A") = - 5 i + 9 j$

The magnitude of a vector is of the form:

$\Rightarrow ∣ ∣ ∣ \to A ∣ ∣ ∣ = \sqrt{x^{2} + y^{2}}$ $Rightarrow |vec("A")| = sqrt(x^(2) + y^(2))$

Substituting the relevant values:

$\Rightarrow ∣ ∣ ∣ \to A ∣ ∣ ∣ = \sqrt{{(- 5)}^{2} + {(9)}^{2}}$ $Rightarrow |vec("A")| = sqrt((- 5)^(2) + (9)^(2))$

$\Rightarrow ∣ ∣ ∣ \to A ∣ ∣ ∣ = \sqrt{25 + 81}$ $Rightarrow |vec("A")| = sqrt(25 + 81)$

$therefore |vec("A")| = sqrt(106)$

Therefore, the magnitude of vector $"A"$ is $sqrt(106)$ .