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`.

Answer 1

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)`