Question

Consider three vectors a=i+j+k,b=i+j-k and c=i-j. If k1a+k2b+k3c=4i+6j+k,where k1,k2,k3 are scalars.then k1=

Answer 1

Answer:

a = i+j+k

b = i+j–k

c = i–j

If K1a + k2b +k3c = 4i +6j +k

k1 = 4

k2 = 2

k3= 1

Hope It Helps You,

Answer 2

Answer:

Required value of k1 is 7/2.

Explanation:

Given,

$a=i+j+k,b=i+j-k \: and \: c=i-j$

And,

$k1a+k2b+k3c=4i+6j+k.......(1)$

Putting value of a,b,c in the equation (1),

$(i+j+k)k1 + (i+j-k)k2 +(i-j)k3 = 4i+6j+k$

$(k1+k2+k3)i+(k1+k2-k3)j+(k1-k2)k=4i+6j+k$

Comparing both side and eliminateliminating i,j,k and get,

$(k1+k2+k3) = 4 ......(1)\\ (k1+k2-k3)=6....(2) \\ (k1-k2)=1.....(3)$

From (3),

$k2 = k1 - 1......(4)$

From equation (2) and (4) and (5),

$k3 = 6 - k1 - k2 = 6 - k1 - k1 + 1 = 7 - 2k1.......(5)$

From equation (1),(2)

$k1 + k1 - 1 - 7 + 2k1 = 6 \\ 4k1 - 8 = 6 \\ 4k1 = 6 + 8 \\ k1 = \frac{14}{4} = \frac{7}{2}$

Value of k1 is 7/2.