Math

QuestionFind the value(s) of hh such that the vector b=[53h]b=\left[\begin{array}{l}5 \\ 3 \\ h\end{array}\right] lies in the plane spanned by a1=[131]a_{1}=\left[\begin{array}{r}1 \\ 3 \\ -1\end{array}\right] and a2=[5112]a_{2}=\left[\begin{array}{r}-5 \\ -11 \\ 2\end{array}\right]. The value(s) of hh is(are) \square.

Studdy Solution

STEP 1

Assumptions1. We have three vectors a1a_{1}, aa_{}, and bb. . The vector bb is in the plane spanned by a1a_{1} and aa_{}.
3. The plane spanned by a1a_{1} and aa_{} is the set of all linear combinations of a1a_{1} and aa_{}.

STEP 2

If bb is in the plane spanned by a1a_{1} and a2a_{2}, then bb can be expressed as a linear combination of a1a_{1} and a2a_{2}. This gives us the equationb=c1a1+c2a2b = c_{1}a_{1} + c_{2}a_{2}where c1c_{1} and c2c_{2} are scalars.

STEP 3

We can write the equation in terms of the components of the vectors[53h]=c1[131]+c2[5112]\left[\begin{array}{l}5 \\3 \\ h\end{array}\right] = c_{1}\left[\begin{array}{r}1 \\3 \\ -1\end{array}\right] + c_{2}\left[\begin{array}{r}-5 \\ -11 \\2\end{array}\right]

STEP 4

This gives us a system of three equations1. =c1c2 = c_{1} -c_{2}
2. 3=3c111c23 =3c_{1} -11c_{2}
3. h=c1+2c2h = -c_{1} +2c_{2}

STEP 5

We can solve the first two equations to find the values of c1c_{1} and c2c_{2}.

STEP 6

Multiply the first equation by3 and the second equation by1, we get1. 15=3c115c215 =3c_{1} -15c_{2}
2. 3=3c111c23 =3c_{1} -11c_{2}

STEP 7

Subtract the second equation from the first to get12=4c212 = -4c_{2}

STEP 8

olve for c2c_{2}c2=3c_{2} = -3

STEP 9

Substitute c2=3c_{2} = -3 into the first equation to solve for cc_{}5=c5(3)5 = c_{} -5(-3)

STEP 10

olve for cc_{}c=515=10c_{} =5 -15 = -10

STEP 11

Substitute c=10c_{} = -10 and c=3c_{} = -3 into the third equation to solve for hhh=(10)+(3)h = -(-10) +(-3)

STEP 12

olve for hhh=106=4h =10 -6 =4The value of hh is4.

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