Let $\overrightarrow a = \hat i + 4\hat j + 2\hat k,\overrightarrow b = 3\hat i - 2\hat j + 7\hat k$ and $\vec c = 2\hat i - \hat j + 4\hat k$. Find a vector $\vec d$ which is perpendicular to both $\vec a$ and $\bar b$, and $\vec c \cdot \vec d = 15 \cdot$

Let $\hat d = x\hat i + y\hat j + z\hat k$ …(i)
Since $\hat d$ is perpendicular to $\widehat a$

, we get
$(x\hat i + y\hat j + z\hat k) \cdot (\hat i + 4\hat j + 2\hat k) = 0$

$\Rightarrow$ $x + 4y + 2z = 0$

…(ii)
and $\hat d$ is perpendicular to $\hat b$ ,
$\therefore$ $(x\hat i + y\hat j + z\hat k) \cdot (3\hat i - 2\hat j + 7\hat k) = 0$
$\Rightarrow 3x - 2y + 7z = 0$

…(iii)
Also, $\vec c \cdot \vec d = 15$
$\Rightarrow (2i - j + 4k) \cdot (xi + y\hat j + z\hat k) = 15$

$\Rightarrow 2x - y + 4z = 15$

…(iv)
(iii)$- 3$ (ii) gives: $- 14y + z = 0 \Rightarrow z = 14y$

…(v)
(iv)$- 2$ (ii) gives $- 9y = 15 \Rightarrow y = - \cfrac{5}{3}$

…(vi)
Putting in (v),

we get $z = - \cfrac{{70}}{3}$

Putting in (ii),

we get $x - \cfrac{{20}}{3} - \cfrac{{140}}{3} = 0 \Rightarrow x = \cfrac{{160}}{3}$

Putting in (i),

we get $\overrightarrow d = \cfrac{{160}}{3}\widehat i - \cfrac{5}{3}\widehat j - \cfrac{{70}}{3}\widehat k = \cfrac{1}{3}\left( {160\widehat i - 5\widehat j - 70\widehat k} \right)$