Question

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

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Answer 1

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

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

Verification:

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

$\sqrt{ {x}^{2} + {y}^{2} }$

Thus,

Here

Magnitude:

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

Final answer:

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

or

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

Hope it helps you.

To learn more on brainly:

Prove that the vectors (2, 1, 4), (1, –1, 2) and (3, 1, –2) form a basis of V3(R)

https://brainly.in/question/41923499