Question

Let a=(1,2,3,4), b=(4,3,2,1) and c=(1,1,1,1) be vectors in R4. Part (a) [4 points]: Find (a⋅2c)b+||−3c||a. Part (b) [6 points]: Find two perpendicular vectors p and q in R4 such that their sum is the vector b and such that p is parallel to a. Part (c) [3 points]: If T(−1,1,2,−2) is the terminal point of the vector a, then what is its initial point? Part (d) [2 points]: Find a vector in R4 that is perpendicular to b.

Answer 1

Given :

a = (1, 2, 3, 4) , b = ( 4, 3, 2, 1), c = (1, 1, 1, 1) ∈
`R^4`

a). (a.2c)b + ||-3c||a

Now,

(a.2c) = (1, 2, 3, 4). 2 (1, 1, 1, 1)

= (2 + 4 + 6 + 6)

= 20

-3c = -3 (1, 1, 1, 1)

= (-3, -3, -3, -3)

||-3c|| =
`$√((-3)^2 + (-3)^2 + (-3)^2 + (-3)^2 )$`

`$=√(9+9+9+9)$`

`$=√(36)$`

= 6

Therefore,

(a.2c)b + ||-3c||a = (20)(4, 3, 2, 1) + 6(1, 2, 3, 4)

= (80, 60, 40, 20) + (6, 12, 18, 24)

= (86, 72, 58, 44)

b). two vectors
`\vec A` and
`\vec B` are parallel to each other if they are scalar multiple of each other.

i.e.,
`\vec A=r \vec B` for the same scalar r.

Given
`\vec p` is parallel to
`\vec a`, for the same scalar r, we have

`$\vec p = r (1,2,3,4)$`

`$\vec p = (r,2r,3r,4r)$` ......(1)

Let
`\vec q = (q_1,q_2,q_3,q_4)` ......(2)

Now given
`\vec p` and
`\vec q` are perpendicular vectors, that is dot product of
`\vec p` and
`\vec q` is zero.

`$q_1r + 2q_2r + 3q_3r + 4q_4r = 0$`

`$q_1 + 2q_2 + 3q_3 + 4q_4 = 0$` .......(3)

Also given the sum of
`\vec p` and
`\vec q` is equal to
`\vec b`. So

`\vec p + \vec q = \vec b`

`$(r,2r,3r,4r) + (q_1+q_2+q_3+q_4)=(4, 3,2,1)$`

∴
`$q_1 = 4-r , \ q_2=3-2r, \ q_3 = 2-3r, \ q_4=1-4r$` ....(4)

Putting the values of
`q_1,q_2,q_3,q_4` in (3),we get

`r=(2)/(3)`

So putting this value of r in (4), we get

`$\vec p =\left( (2)/(3), (4)/(3), 2, (8)/(3) \right)$`

`$\vec q =\left( (10)/(3), (5)/(3), 0, (-5)/(3) \right)$`

These two vectors are perpendicular and satisfies the given condition.

c). Given terminal point is
`\vec a` is (-1, 1, 2, -2)

We know that,

Position vector = terminal point - initial point

Initial point = terminal point - position point

= (-1, 1, 2, -2) - (1, 2, 3, 4)

= (-2, -1, -1, -6)

d).
`\vec b = (4,3,2,1)`

Let us say a vector
`\vec d = (d_1, d_2,d_3,d_4)` is perpendicular to
`\vec b.`

Then,
`\vec b.\vec d = 0`

`$4d_1+3d_2+2d_3+d_4=0$`

`$d_4=-4d_1-3d_2-2d_3$`

There are infinitely many vectors which satisfies this condition.

Let us choose arbitrary
`$d_1=1, d_2=1, d_3=2$`

Therefore,
`$d_4=-4(-1)-3(1)-2(2)$`

= -3

The vector is (-1, 1, 2, -3) perpendicular to given
`\vec b.`