Question

Let A=3i+4j,B=2i+2j−k, and C=3i−4k.

(a) Find ∣A∣,∣B∣, and ∣C∣.
(b) For what values of s is ∣sB∣=1.
(c) Find the unit vector having the same direction as A.

Answer 1

The magnitudes of the vectors A, B, and C are 5, 3, and 5 respectively. The value of s such that |sB| equals 1 is 1/3. The unit vector with the same direction as A is (3/5)i + (4/5)j.

Step-by-step explanation:

The magnitude of the vectors A=3i+4j, B=2i+2j−k, and C=3i−4k are found using the formula for the magnitude of a vector: |V| = √(v_x^2 + v_y^2 + v_z^2).

(a) To find the magnitudes |A|, |B|, and |C|, we compute:

• |A| = √(3^2 + 4^2) = 5
• |B| = √(2^2 + 2^2 + (-1)^2) = 3
• |C| = √(3^2 + (-4)^2) = 5

(b) To find the values of s such that |sB| = 1, we use the magnitude of B found in part (a) and solve:

s*|B| = 1

s*3 = 1

s = 1/3

(c) To find the unit vector having the same direction as A, we divide each component of A by its magnitude:

Unit vector of A = (1/|A|)*A = (1/5)*(3i+4j) = (3/5)i + (4/5)j