A vector A is added to B=6i-8j. The resultant vector is in the positive x direction and has a magnitude equal to A . What is the

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A vector A is added to B=6i-8j. The resultant vector is in the positive x direction and has a magnitude equal to A . What is the

magnitude of A?
a)11
b)7.1
c)5.1
d)8.3
e)12.2

Physics

2 answers:

Ostrovityanka [3.2K] · Answer 1 · 2026-03-02T16:58:04+03:00

The value of $\vec A$ is $8.3$ .

Additional details:

A vector is a measurement defined by both magnitude and direction. It can be depicted as the product of its magnitude and the corresponding direction vector.

Information provided:

We have the vector as $\vec B = 6\hat i - 8\hat j$ .

The resultant vector is oriented towards the X-direction.

Concept applied:

Take the vector $\vec A = a\hat i + b\hat j$ , which is combined with $\vec B$ to yield the resultant vector $\vec C$ . The resultant vector moves in the positive X-direction, indicating that its Y-component is zero.

The formula for the resultant vector is expressed as:

$\vec C = \vec A + \vec B$

Replace $6\hat i - 8\hat j$ with $\vec B$ in this equation.

$\begin{gathered}\vec C = \left( {a\hat i + b\hat j} \right) + \left( {6\hat i - 8\hat j} \right) \\= \left( {a + 6} \right)\hat i + \left( {b - 8} \right)\hat j \\ \end{gathered}$

The resultant vector can be denoted as:

$\vec C = x\hat i + 0\hat j$

Evaluate the two expressions of the resultant vector.

$x = a + 6$ …… (1)

$\begin{aligned}0&=b-8\\b&=8\\\end{aligned}$

The magnitude of the resultant vector corresponds to the magnitude of $\vec A$ .

The expression for the magnitude of $\vec A$ is given by:

$\left| {\vec A} \right| = \sqrt {\left( {{a^2}} \right) + \left( {{b^2}} \right)}$ …… (2)

The expression for the magnitude of $\vec C$ is written as:

$\left| {\vec C} \right| = \sqrt {{{\left( {a + 6} \right)}^2}}$

Compare the two previous expressions.

$\sqrt {\left( {{a^2}} \right) + \left( {{b^2}} \right)}= \sqrt {{{\left( {a + 6} \right)}^2}}$

Insert 8 for $b$ into the preceding expression.

$\begin{aligned}\sqrt {\left( {{a^2}} \right) + \left( {{8^2}} \right)}&=\sqrt {{{\left( {a + 6} \right)}^2}}\\{a^2} + 64&={a^2} + 36 + 12a \\12a&=28 \\a&=2.33 \\ \end{aligned}$

Use 2.33 for 'a' and 8 for 'b' in equation (2).

$\begin{aligned}\left| {\vec A} \right|&=\sqrt {{{\left( {2.33} \right)}^2} + {{\left( 8 \right)}^2}}\\&=8.33 \\ \end{aligned}$

Therefore, the magnitude of vector A turns out to be $8.33$ .

Learn more:

1. Motion under friction .

2. Conservation of momentum .

3. Circular motion .

Answer Details:

Grade: College

Subject: Physics

Chapter: Vectors

Keywords:

Vectors, product, magnitude, direction, resultant vector, adding vector, subtraction of vector, 2.33, 8, 8.33, 8.3, 8.33.

Yuliya22 [3.3K] · Answer 2 · 2026-03-02T17:00:00+03:00

The answer is letter d, 8.3.

Here’s a solution for the given problem:

We have:

B = 6i - 8j

Let A be unknown; we'll denote A as = mi + nj

The resultant A+B lies along the x-axis (which implies A+B = Ki + 0j, where K is yet to be determined,

and we also know the magnitude of A+B is equivalent to the magnitude of A,

therefore, mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2.

Using vector addition, A+B becomes (m+6)i + (n-8)j.

Since we know A+B = Ki + 0j, we can establish that:

m + 6 = K

n - 8 = 0, which gives n=8.

Thus, K^2=m^2+n^2 means (m+6)^2 = m^2 +8^2

= m^2 + 12m + 36 = m^2 + 64

which gives us 12m = 28

m = 2.33333...

Consequently, the magnitude of A is sqrt[(2.333...)^2 + 8^2] = 8.3333.