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Ksju [112]
3 years ago
8

Find the projection of the vector A = î - 2ġ + k on the vector B = 4 i - 4ſ + 7k. 15. Given the vectors A = 2 i +3 ſ +6k and B =

i +59 +3k. How much of vector B is along vector A?
Mathematics
1 answer:
Gwar [14]3 years ago
7 0

Answer:

Part 1)

Projection of vector A on vector B equals 19 units

Part 2)

Projection of vector B' on vector A' equals 35 units

Step-by-step explanation:

For 2 vectors A and B the projection of A on B is given by the vector dot product of vector A and B

Given

\overrightarrow{v_{a}}=\widehat{i}-2\widehat{j}+\widehat{k}

Similarly vector B is written as

\overrightarrow{v_{b}}=4\widehat{i}-4\widehat{j}+7\widehat{k}

Thus the vector dot product of the 2 vectors is obtained as

\overrightarrow{v_{a}}\cdot \overrightarrow{v_{b}}=(\widehat{i}-2\widehat{j}+\widehat{k})\cdot (4\widehat{i}-4\widehat{j}+7\widehat{k})\\\\\overrightarrow{v_{a}}\cdot \overrightarrow{v_{b}}=1\cdot 4+2\cdot 4+1\cdot 7=19

Part 2)

Given vector A' as

\overrightarrow{v_{a'}}=2\widehat{i}+3\widehat{j}+6\widehat{k}

Similarly vector B' is written as

\overrightarrow{v_{b'}}=\widehat{i}+5\widehat{j}+3\widehat{k}

Thus the vector dot product of the 2 vectors is obtained as

\overrightarrow{v_{b'}}\cdot \overrightarrow{v_{a'}}=(\widehat{i}+5\widehat{j}+3\widehat{k})\cdot (2\widehat{i}+3\widehat{j}+6\widehat{k})\\\\\overrightarrow{v_{a'}}\cdot \overrightarrow{v_{b'}}=1\cdot 2+5\cdot 3+3\cdot 6=35

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