Question

magnitude of vector 3i-12j-4k

Answer 1

Its underroot of 144 +16+9
Ans is underroot 169...that is 13

Magnitide is 13

Answer 2

Answer:

The magnitude of the vector 3i-12j-4k is 13.

Explanation:

Vector: A vector is a physical quantity that has a magnitude and direction. It is represented by a line segment with an arrow. It is denoted as variable with an arrow on top of it.

Examples of vector quantities: displacement, velocity, Force etc.

Magnitude of a vector: The magnitude of the vector represents the value of the physical property it represents. It is represented by the length of the vector. It is denoted by |a|.

Direction of a vector: It is the angle the vector makes with the X, Y, Z - axes. It represents the direction of the physical quantity. It is in the direction of from tail to the arrow head.

Unit vectors i, j, k: i, j, k are the vectors with unit length with directions coinciding with the X, Y, Z axes respectively. These unit vectors are used to represent the direction of any vector in terms of X, Y, Z axes.

Calculation:

Magnitude of a vector p = ai +bj + ck is calculated as |p| = $\sqrt{a^{2} +b^{2} +c^{2} }$

The magnitude of 3i-12j-4k is

$\sqrt{3^{2} +(-12)^{2} +(-4)^{2} } \\\\=\sqrt{9+144+16} \\\\=\sqrt{ 169}\\\\=13$

The magnitude of 3i-12j-4k is 13.

Know more about vector operations:

Resolution of vectors

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Vector product and its properties

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