Question

given A=4i-10j and B= 7i+5j find b such that A+bB is a vector pointing along the x-axis (i.e has no y component)​

Answer 1

The value of $\beta$ such that $\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}$ is 2.

According to the statement, we have following system of vectorial equations:

$\vec A = 4\,\hat {i} - 10\,\hat{j}$ (1)

$\vec {B} = 7\,\hat{i} + 5\,\hat{j}$ (2)

$\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}$ (3)

By applying (1) and (2) in (3):

$(4\,\hat{i}-10\,\hat{j}) + \beta\cdot (7\,\hat{i}+5\,\hat{j}) = c\,\hat{i}$

$(4+7\cdot \beta)\,\hat{i} +(-10+5\cdot \beta)\,\hat{j} = c\,\hat{i}$

And we get two scalar equations after analyzing each component:

$4+7\cdot \beta = c$ (4)

$-10+5\cdot \beta = 0$ (5)

We solve for $\beta$ in (5):

$\beta = 2$

And for $c$ in (4):

$c = 4+7\cdot (2)$

$c = 18$

The value of $\beta$ such that $\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}$ is 2.

Answer 2

Answer:

-4/7

Explanation:

Given the following

A=4i-10j and B= 7i+5j

A+ bB = 4i-10j + (7i+5j)b

A+ bB =  4i-10j + 7ib+5jb

A+ bB =

The vector along the x-axis is expressed as i + 0j

If the vector A+ bB is pointing in the direction of the x-axis then;

$A+ bB * \frac{i+0j}{|i+0j|} = 0 \\ (4+7b)i-(10-5b)j* \frac{i+0j}{\sqrt{1^2+0^2} } = 0\\(4+7b)i-(10-5b)j *(i+0j) = 0\\4+7b-0 =0\\7b=-4\\b = -4/7$

Hence the value of b is -4/7