Question

QQ.A= 4i - 3j , B = 6i - 5j . the vector having magnitude 7 parallel to B - A is

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Explanation:

Given: A= 4i - 3j , B = 6i - 5j .

To find: The vector having magnitude 7 parallel to B - A is ?

Solution:

Step 1: Find B-A

B-A = 6i-5j-4i+3j

B-A = 2i-2j

Step 2: The vector parallel to B-A is =k(2i-2j).

Put the value of k so that the magnitude of vector is 7.

k should be 7/2√2

Thus, the vector which is parallel to B-A is

$\begin{gathered} \frac{7}{2 \sqrt{2} } (2i - 2j) \\ \end{gathered}$

Verification:

If a vector is xi+yj then it's Magnitude is

$\begin{gathered} \sqrt{ {x}^{2} + {y}^{2} } \\ \end{gathered}$

Thus,

Here

Magnitude:

$\begin{gathered} = \sqrt{ ({ \frac{14}{2 \sqrt{2} } })^{2} + ({ \frac{ - 14}{2 \sqrt{2} } })^{2} } \\ \\ = \sqrt{ ({ \frac{7}{\sqrt{2} } })^{2} + ({ \frac{ - 7}{ \sqrt{2} } })^{2} } \\ \\ = \sqrt{ { \frac{49}{2} } + \frac{49}{2} } \\ \\ = \sqrt{ \frac{49 + 49}{2} } \\ \\ \sqrt{ \frac{98}{2} } \\ \\ = \sqrt{49} \\ \\ = 7\end{gathered}$

Final answer:

$\begin{gathered}\frac{7}{2 \sqrt{2} } (2i - 2j) \\ \\ \end{gathered}$

or

$\begin{gathered} \frac{7}{ \sqrt{2} } (i - j) \\ \\ \end{gathered}$

Hope it helps you.

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