Question

Let i, j and k be unit vectors in the x, y and z directions, respectively. Suppose that a = 3i − j + 2k. What is the magnitude of the vector a?

Answer 1

Answer:

Example 1 Find the unit vector in the direction of the sum of the vectors a.. = ˆ. ˆ ˆ. 2. 2. i j k. − + and b.. = –. ˆ. ˆ ˆ 3. i j k.

Explanation:

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Answer 2

Answer:

The magnitude of the vector 'a' is √(14).

Explanation:

Given,

i, j, and k are the unit vectors along the directions x, y, and z directions.

The vector a = 3i - j + 2k

To find,

The magnitude of the vector a = 3i - j + 2k.

Concept,

Let A be a vector with x units along the x-direction, y units along the y-direction, and z units along the z-direction.

i.e. A = xi + yj + zk then, the magnitude of the vector A is given by:
|A| = $\sqrt{x^2 + y^2 + z^2}$....(1)

Calculation,

Now here the vector A = a, x = 3, y = -1, z = 2 substituting in the equation (1):
|A| = $\sqrt{3^2 + (-1)^2 + 2^2}$

⇒ |A| = $\sqrt{9 + 1 + 4}$

⇒ |A| = √(14)

Therefore, the magnitude of the vector 'a' is √(14).

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