Question

The norm of the vector (1,-2,-3) is

Answer 1

Answer:

sqrt{14}

Explanation:

We recall the concept of norm of a vector

Norm of a vector is length of the vector

Given:

Vector is (1,-2,-3)

Norm of a vector =\sqrt{1^{2}+(-2)^{2} +(-3)^{2}  }

                           =\sqrt{1+4+9}

                           =\sqrt{14}

Answer 2

Answer:

The norm of given vector is

$\sqrt{14} \: units$

Explanation:

Given that the point in cartesian coordinate system is (1,-2,-3)

We know that for a point (a,b,c) the position vector of the point is

$r = ai + bj + ck$

Hence,the position vector of given point is as follows

$r = i - 2j - 3k$

Norm of vector is nothing but the magnitude or length of vector.Hence,the norm is

$|r| = \sqrt{1 {}^{2} + ( - 2) {}^{2} + ( - 3) {}^{2} } \\ = \sqrt{14} \: units$

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