Question

Motion in a Plane
and difference of two vectors A and B are
The sum and diff
B-2 . +6ộ + k and à - B = 4 î+25 - 112.
magnitude of each vector and their scalar
Find the magnitude of en
product AB.
Ans. V50, 41, - 25
​

Answer 1

Explanation:

Given: A + B = 2i + 6j - k

Given: A - B = 4i + 2j - 11k

On Solving (i) & (ii), we get

A + B = 2i + 6j + k

A - B = 4i + 2j - 11k

------------------------------

2A = 6i + 8j - 10k

A = 3i + 4j - 5k

(i)

Magnitude of A = √(3)² + (4)² - (5)²

                          = √50

Substitute A = 3i + 4j - 5k in above equations, we get

⇒ 3i + 4j - 5k + B = 2i + 6j + k

⇒ B = -i + 2j + 6k

(ii)

Magnitude of B = √(-1)² + (2)² + (6)²

                          = √1 + 4 + 36

                          = √41

(iii)

Dot product (A.B) = 3 * (-1) + 4 * 2 + (-5) * 6

                             = -3 + 8 - 30

                             = -25

Hope it helps!

Answer 2

Let A=x i^ +y j^+z k^, B=a i^+b j^+ c k^.[i^ -->i cap, j^ -->j cap, k^ -->k cap].  

Therefore, x+a=2.....(1) , x-a=4....(2).  

Adding (1)&(2),  

2x=6  

--> x=3.  

So, a = 2-x = 2-3 = -1.  

Similarly, y+b=6.....(3) , y-b=2.....(4)  

Adding (3)&(4),  

2y=8  

--> y=4.  

So, b = 6-y = 6-4 = 2.  

Finally, z+c=1....(5) , z-c= -11.....(6)  

Adding (5)&(6),  

2z=-10  

-->z=-5  

So, c = 1-z = 1-(-5) = 6.  

Thus, A = 3 i^ + 4 j^ - 5 k^, B = -1 i^ + 2 j^ + 6 k^.  

Therefore = |A| = sq rt(x^2+y^2+z^2) = sq rt(9+16+25) = sq rt(50).[|A|-->Magnitude of vector A]  

|B| = sq rt(1+4+36) = sq rt(41) .  

Other than the form u gave, A.B is also represented by [(x*a)+(y*b)+(z*c)]. So, A.B here is equal to (3*[-1])+(4*2)+([-5]*6) = -3+8-30 = -25.