Question

2i+k=a, 3j +4K=b, 8i-3j=c,
if a =xb+yc then (x, y) =​

Answer 1

Answer:

Step-by-step explanation:

2i+k=a, 3j +4K=b, 8i-3j=c

and a =xb+yc

so putting the values of a, b and c in the above equation we get,

2i+k = x(3j +4K) + y(8i-3j)

=> 2i+k = 3xj+4xk+8yi-3yj

=> 2i+k = 8yi + (3x-3y)j + 4xk

Now equating both side we get,

(2)i = (8y)i

=> 2 = 8y --------------- (1)

=> y = 1/4

(1)k = (4x)k

=> 1 = 4x ---------------- (2)

=> x = 1/4

So the valu 0f (x,y) = (1/4 , 1/4)

Hope it helps you.

Answer 2

Values of (x,y) are $\bf \left( \frac{1}{4}, \frac{1}{4} \right)$.

Given:

• $2i + k = a$
• $3j + 4k = b$and
• $8i - 3j = c$

To find:

• Find the value of (x,y), if $a = xb + yc$

Solution:

Step 1:

Put the given values in the condition.

As,

$2i+k=a, 3j +4k=b, 8i-3j=c$

So,

$2i+k=x( 3j +4k) + y(8i-3j)$

$\bf 2i+k=3jx +4kx + 8iy-3jy$

Step 2:

Compare the similar terms.

$2i+k=8iy +( 3x - 3y)j+4kx$

Compare the coefficients of i, j and k.

Compare the coefficient of i:

$2 = 8y$

$y = \frac{2}{8}$

$\bf y = \frac{1}{4}$

Compare the coefficient of k:

$4x = 1$

$\bf x = \frac{1}{4}$

Thus,

Values of (x,y) are $\bf \left( \frac{1}{4}, \frac{1}{4} \right)$.

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