Question

If the resultant of the vectors (i+2j-k). (i-j+2k) and
is a unit vector along the y-direction, then C is
(2) -2i-k
(c) 2i-k
(b) –2i+k
(d) 2i+k​

Answer 1

Explanation:

Let A = i+2j -k

B = I - j +2k

A = √ 1+4+1 = √6

B = √1 +1+4 = √6

C is resultant vector......

It's along Y axis so theta = 90°

C = √A^2 + B^2 + 2AB Cos theta

C = √ 6 + 6 + 2√36 (1)

C^2 = 12 + √4×36

C^2 = 12 + 12

C = √ 24

C = 2 √6

Answer 2

Concept:

We need to apply the concept of the sum of the total vector. A + B + C = unit vector along y direction.

Given:

Vector A = (i + 2j - k)

Vector B = (i - j + 2k)

Resultant of vector = unit vector along the y-direction

Find:

We need to determine the vector C from the resultant of the vectors A, B and C.

Solution:

Let vector C be (xi + yj + zk)

The resultant vector is a unit vector.

Therefore sum of vector A + B + C = 1j

We have vector A and vector B as (i + 2j - k) and (i - j + 2k) respectively.

Therefore, equation becomes-

i + 2j - k + i - j + 2k + xi + yj + zk = 1j

(2 + x)i + ( 1 + y)j + ( 1 + z)k = 1j

Therefore, equating components from L.H.S to R.H.S  becomes,

2+x = 0

x = -2

1 + y = 1

y = 0

1 + z = 0

z = - 1

Therefore, vector C becomes =  -2i - k

Thus, the vector C is -2i - k

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