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There are two vectors vecA= 3hati+hatj ...

There are two vectors `vecA= 3hati+hatj and vecB=hatj+2hatk`. For these two vectors -
(a) If `vecA & vecB` are the adjacent sides of a parallelogram then find the magnitude of its area.
(b) Find a unit vector which is perpendicular to both `vecA & vecB`.

Text Solution

Verified by Experts

The correct Answer is:
(a) 7 units
(b) `(2)/(7)hati- (6)/(7)hatj+ (3)/(7)hatk`

(a) Area of the parallelogram `= |vecA xx vecB| = |{:(hati,,hatj,,hatk),(3,,1,,0),(0,,1,,2):}|`
`" "= |2hati-6hatj+3hatk|= sqrt(2^(2)+ (-6)^(2)+ 3^(2))` = 7 units
(b) Unit vector perpendicular to both `vecA & vecB`
`hatn = (vecAxx vecB)/(|vecAxx vecB|)= (2hati-6hatj+3hatk)/(7)`
`= (2)/(7)hati- (6)/(7)hatj+ (3)/(7) hatk`
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Knowledge Check

  • If vecA = 3hati + 4hatj and vecB = hati + hatj + 2hatk then find out unit vector along vecA + vecB .

    A
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  • If vecA= 3hati+2hatj and vecB= 2hati+ 3hatj-hatk , then find a unit vector along (vecA-vecB) .

    A
    `(hati-hatj+hatk)/(sqrt3) `
    B
    `(hati-hatj+hatk)/(sqrt5) `
    C
    `(hati+hatj+hatk)/(sqrt3) `
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  • Given : vecA = 2hati - hatj + 2hatk and vecB = -hati - hatj + hatk . The unit vector of vecA - vecB is

    A
    `(3hati + hatk)/(sqrt10)`
    B
    `(3hati)/(sqrt10)`
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