Respuesta :
The correct answer would be letter d, 8.3.
Solution for the problem follows:
Given are:
B = 6i - 8j
A is unknown; let A be = mi + nj
A+B is along the x axis (therefore A+B = Ki + 0j, where K is unknown,
but then again the magnitude of A+B is the similar as the magnitude of A,
so mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2.
A+B, from simple vector addition, will be now (m+6)i + (n-8)j.
Ever since we previously
know A+B = Ki + 0j, we now know that:
m+6 = K
n-8 = 0, which implies n=8.
Thus, K^2=m^2+n^2 ====> (m+6)^2 = m^2 +8^2
= m^2 + 12m + 36 = m^2 + 64
= 12m = 28
= m = 2.33333...
Therefore, the magnitude of A is sqrt[(2.333...)^2 + 8^2] = 8.3333
The magnitude of [tex]\vec A[/tex] is [tex]8.3[/tex].
Further explanation:
A vector is a quantity having magnitude and direction both. It is represented as the product of magnitude and direction vector.
Given:
The vector is given as [tex]\vec B = 6\hat i - 8\hat j[/tex].
The direction of resultant vector is in the X- direction.
Concept used:
Consider a vector [tex]\vec A = a\hat i + b\hat j[/tex] which is added to [tex]\vec B[/tex] to get a resultant vector [tex]\vec C[/tex]. The resultant vector is directed in the positive X-direction which means that the y- component of the vector is zero.
The expression for the resultant vector is given as.
[tex]\vec C = \vec A + \vec B[/tex]
Substitute [tex]6\hat i - 8\hat j[/tex] for [tex]\vec B[/tex] in the above expression.
[tex]\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}[/tex]
The resultant vector is represented as.
[tex]\vec C = x\hat i + 0\hat j[/tex]
Compare the above two expression of resultant vector.
[tex]x = a + 6[/tex] …… (1)
[tex]\begin{aligned}0&=b-8\\b&=8\\\end{aligned}[/tex]
The magnitude of resultant vector is equal to the magnitude of [tex]\vec A[/tex].
The expression for the magnitude of [tex]\vec A[/tex] is given as.
[tex]\left| {\vec A} \right| = \sqrt {\left( {{a^2}} \right) + \left( {{b^2}} \right)}[/tex] …… (2)
The expression for the magnitude of [tex]\vec C[/tex] is given as.
[tex]\left| {\vec C} \right| = \sqrt {{{\left( {a + 6} \right)}^2}}[/tex]
Compare the above two expressions.
[tex]\sqrt {\left( {{a^2}} \right) + \left( {{b^2}} \right)}= \sqrt {{{\left( {a + 6} \right)}^2}}[/tex]
Substitute 8 for [tex]b[/tex] in above expression.
[tex]\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}[/tex]
Substitute 2.33 for a and 8 for b in equation (2).
[tex]\begin{aligned}\left| {\vec A} \right|&=\sqrt {{{\left( {2.33} \right)}^2} + {{\left( 8 \right)}^2}}\\&=8.33 \\ \end{aligned}[/tex]
Thus, the magnitude of vector A is [tex]8.33[/tex].
Learn more:
1. Motion under friction https://brainly.com/question/7031524.
2. Conservation of momentum https://brainly.com/question/9484203.
3. Circular motion https://brainly.com/question/9575487.
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.
