We read your photo as:

Given that P = (-4/5), q = (1/2) and R = (2/5), find: (i) P + 2R (ii) 3q - P (iii) 2P - q + 3R

Mathematics

ScanSolve AI Answer

Answers

3 parts
i

(015)\begin{pmatrix} 0 \\ 15 \end{pmatrix}

ii

(71)\begin{pmatrix} 7 \\ 1 \end{pmatrix}

iii

(−323)\begin{pmatrix} -3 \\ 23 \end{pmatrix}

Step-by-step solution

9 steps

Here are the solutions for the given vector operations.

Given vectors:

P=(−45),Q=(12),R=(25)P = \begin{pmatrix} -4 \\ 5 \end{pmatrix}, \quad Q = \begin{pmatrix} 1 \\ 2 \end{pmatrix}, \quad R = \begin{pmatrix} 2 \\ 5 \end{pmatrix}

Part i· 3 steps

Find P+2RP + 2R

  1. Substitute the given vectors into the expression.

    P+2R=(−45)+2(25)P + 2R = \begin{pmatrix} -4 \\ 5 \end{pmatrix} + 2 \begin{pmatrix} 2 \\ 5 \end{pmatrix}

  2. Perform the scalar multiplication.

    P+2R=(−45)+(2×22×5)=(−45)+(410)P + 2R = \begin{pmatrix} -4 \\ 5 \end{pmatrix} + \begin{pmatrix} 2 \times 2 \\ 2 \times 5 \end{pmatrix} = \begin{pmatrix} -4 \\ 5 \end{pmatrix} + \begin{pmatrix} 4 \\ 10 \end{pmatrix}

  3. Add the resulting vectors.

    P+2R=(−4+45+10)=(015)P + 2R = \begin{pmatrix} -4 + 4 \\ 5 + 10 \end{pmatrix} = \begin{pmatrix} 0 \\ 15 \end{pmatrix}

    The answer for (i) is (015)\boxed{\begin{pmatrix} 0 \\ 15 \end{pmatrix}}.

Part ii

(71)\begin{pmatrix} 7 \\ 1 \end{pmatrix}

Find 3Q−P3Q - P

  1. Substitute the given vectors into the expression.

    3Q−P=3(12)−(−45)3Q - P = 3 \begin{pmatrix} 1 \\ 2 \end{pmatrix} - \begin{pmatrix} -4 \\ 5 \end{pmatrix}

  2. Perform the scalar multiplication.

    3Q−P=(3×13×2)−(−45)=(36)−(−45)3Q - P = \begin{pmatrix} 3 \times 1 \\ 3 \times 2 \end{pmatrix} - \begin{pmatrix} -4 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 \\ 6 \end{pmatrix} - \begin{pmatrix} -4 \\ 5 \end{pmatrix}

  3. Perform the vector subtraction.

    3Q−P=(3−(−4)6−5)=(3+41)=(71)3Q - P = \begin{pmatrix} 3 - (-4) \\ 6 - 5 \end{pmatrix} = \begin{pmatrix} 3 + 4 \\ 1 \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \end{pmatrix}

    The answer for (ii) is (71)\boxed{\begin{pmatrix} 7 \\ 1 \end{pmatrix}}.

Part iii

(−323)\begin{pmatrix} -3 \\ 23 \end{pmatrix}

Find 2P−Q+3R2P - Q + 3R

  1. Substitute the given vectors into the expression.

    2P−Q+3R=2(−45)−(12)+3(25)2P - Q + 3R = 2 \begin{pmatrix} -4 \\ 5 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} + 3 \begin{pmatrix} 2 \\ 5 \end{pmatrix}

  2. Perform the scalar multiplications.

    2P−Q+3R=(2×(−4)2×5)−(12)+(3×23×5)2P - Q + 3R = \begin{pmatrix} 2 \times (-4) \\ 2 \times 5 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} 3 \times 2 \\ 3 \times 5 \end{pmatrix} 2P−Q+3R=(−810)−(12)+(615)2P - Q + 3R = \begin{pmatrix} -8 \\ 10 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} 6 \\ 15 \end{pmatrix}

  3. Perform the vector addition and subtraction.

    2P−Q+3R=(−8−1+610−2+15)2P - Q + 3R = \begin{pmatrix} -8 - 1 + 6 \\ 10 - 2 + 15 \end{pmatrix} 2P−Q+3R=(−9+68+15)=(−323)2P - Q + 3R = \begin{pmatrix} -9 + 6 \\ 8 + 15 \end{pmatrix} = \begin{pmatrix} -3 \\ 23 \end{pmatrix}

    The answer for (iii) is (−323)\boxed{\begin{pmatrix} -3 \\ 23 \end{pmatrix}}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QGiven that P = [[2, 1], [5, −3]] and Q = [[4, −8], [1, −2]], find (2P − Q).Q(3) Given that P-> = (2/3) and Q-> = - (1/4), find the coordinates of PQIv. (2p + 3q − r)(5p − 4r)QFor the points P(−2, 1) and Q(3, 5), in terms of i and j find: a) PQ, b) |PQ|, c) PQQGiven p = 5 −7 , q = −1 3 , and r = 2 1 .
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.