a=2i−3j+k, b=2i−4j+5k, c=− i−4j+2k find the values of p and q if a+pb+qc is parallel to the x axis

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to replace the given vectors, `bar a, bar b, bar c,` in the equation `a + p bar b + q bar c` , such that:

`a + p bar b + q bar c = 2bar i - 3bar j + bar k + p*(2bar i-4bar j+5bar k) + q*(bar i - 4 bar j + 2bar k)`

You need to group the terms containing `bar i, bar j and bar k` , such that:

`a + p bar b + qbar c = (2bar i + 2p*bar i + q*bar i) + ( - 3bar j - 4p*bar j - 4q*bar j) + (bar k + 5p*bar k + 2q*bar k)`

Factoring out `bar i, bar j and bar k,` yields:

`a + p bar b + qbar c = (2 + 2p + q)bar i + (-3 - 4p - 4q)bar j + (1 + 5p + 2q)bar k`

Since `a + p bar b + qbar c` is parallel to x axis, then the coefficients of vectors `bar j` and `bar k` are equal to 0, such that:

`-3 - 4p - 4q = 0 => 4p + 4q = -3`

`1 + 5p + 2q = 0 => 5p + 2q = -1`

Solving the simultaneous equations, yields:

`4p + 4q - 10p - 4q = -3 + 2`

`-6p = -1 => p = 1/6`

Replacing` 1/6` for p in `4p + 4q = -3` , yields:

`4*(1/6) + 4q = -3 => 4 + 24q = -18 => 24q = -22 => q = -22/24 => q = -11/12`

Hence, evaluating p and q, under the given conditions, yields `p = 1/6, q = -11/12.`

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kspcr111 | In Training Educator

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Given ,

`a=2i-3j+k,`

 `b=2i-4j+5k,`  

`c=-i-4j+2k`

As, `a+pb+qc`  is parallel to the x axis first find the a+pb+qc

so,

`a+pb+qc = (2i-3j+k)+p(2i-4j+5k)+q(-i-4j+2k)`


=` 2i-3j+k+2p*i-4pj+5pk-qi-4qj+2qk`

=` (2+2p-q)i+(-3-4p-4q)j+(1+5p+2q)k`

as this vector is parallel to  x-axis, then the coefficient of j and k should be zero. 

so,

`-3-4p-4q=0 and 1+5p+2q=0` .

`-4p-4q =3`

=>`4p+4q =-3`

=> `2p +2q = -3/2` ----------------(1) 

similarly

`1+5p+2q=0`
=>`5p +2q= -1` ----------------------- (2) 

Solving the above equations (1) and (2) we get

`2p +2q = -3/2`

`5p +2q= -1`

`(-) (-) (+)` -----------------on subtracting we get

-------------------

`-3p +0 = -3/2 +1`

=> `-3p = -1/2`

=> `p = 1/6`

so substituting the value of p in the second equation we get

`5p +2q= -1`

`5(1/6) +2q = -1`

=> `5/6 +2q = -1`

=> `2q = -1 -5/6`

=> `2q = -11/6`

=> `q = -11/12`

so,
`p=1/6 and q= - 11/12`

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