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Question
Let `veca = hati + 4hatj + 2hatk, vecb = 3hati - 2hatj + 7hatk ` and `vecc = 2hati - hatj + 4hatk`. Find a vector `vecd` which is perpendicular to both `veca` and `vecb`, and `vecc.vecd = 15`.
Solution
Let `vecd = x hati + y hatj + zhatk` ....(i)
Since `vecd` is perpendicular to `veca,` we get
∴ `(xhati + yhatj + zhatk)* (hati + 4hatj + 2 hatk) = 0`
⇒ x + 4y + 2z = 0 ....(ii)
and `vecd` is perpemdicular to `vecb`
∴ `(xhati + yhatj + zhatk)* (3 hati - 2hatj + 7hatk) = 0`
⇒ 3x - 2y + 7z = 0 ....(iii)
Also `vecc * vecd = 15`
⇒ `(2hati - hatj + 4hatk)* (xhati + yhatj + zhatk) = 15`
⇒ 2x - y + 4z = 15 .....(iv)
(iii) - 3 (ii) gives - 14y + z = 0 ...(v)
(iv) - 2 (ii) gives, -9y = 15 ....(vi)
From (vi), we have `y = -5/3`
Putting in (v), we get `- 14 ((-5)/3) + z = 0`
⇒ `z = -70/3`
Putting in (ii), we get `x - 20/3 - 140/3 = 0`
⇒ `x = 160/3`
Putting in (i), we get
`vecd = 160/3 hati - 5/3 hatj - 70/3 hatk`
`= 1/3 (160 hati - 5hatj - 70 hatk)`
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