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3 years ago

15

3.4=21 2.5=7. 4.4=؟

1 answer:

Licemer1 [7]3 years ago

4 0

Answer:

32

Step-by-step explanation:

From the given values

The possible logic could be

if x.y is given

then it is equal to $=\frac{(x+y)\times x}{y-3}$

For 1st example

3.4 $=\frac{(3+4) \times 3}{4-3} =\frac{21}{1} =21$

For 2nd example

2.5 $=\frac{(2+5) \times 2}{5-3} =\frac{7 \times 2}{2} =7$

⇒4.4 $=\frac{(4+4)\times 4}{4-3} =32$

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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]:

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Solution :

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|| = $\sqrt{(-3)^2 + (-3)^2 + (-3)^2 + (-3)^2 }$

$=\sqrt{9+9+9+9}$

$=\sqrt{36}$

= 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)

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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

$r=\frac{2}{3}$

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

$\vec p =\left( \frac{2}{3}, \frac{4}{3}, 2, \frac{8}{3} \right)$

$\vec q =\left( \frac{10}{3}, \frac{5}{3}, 0, \frac{-5}{3} \right)$

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 $\vec b.$

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