Q.Let $\vec{a} = 4\hat{i} + 5\hat{j} - \hat{k}$, $\vec{b} = \hat{i} - 4\hat{j} + 5\hat{k}$ and $\vec{c} = 3\hat{i} + \hat{j} - \hat{k}$. Find a vector $\vec{d}$ which is perpendicular to both $\vec{c}$ and $\vec{b}$ and $\vec{d} \cdot \vec{a} = 21$.

Since $\vec d$ is perpendicular to both $\vec a$ and $\vec b$, it is parallel to $\vec a\times\vec b$; scale using $\vec c\cdot\vec d=15$.

$\vec a=\hat i+4\hat j+2\hat k$, $\vec b=3\hat i-2\hat j+7\hat k$. Compute $\vec a\times\vec b$:

$\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\1&4&2\\3&-2&7\end{vmatrix}=\hat i(4\cdot7-2\cdot(-2))-\hat j(1\cdot7-2\cdot3)+\hat k(1\cdot(-2)-4\cdot3)$

$=\hat i(28+4)-\hat j(7-6)+\hat k(-2-12)=32\hat i-\hat j-14\hat k.$

Since $\vec d\perp\vec a$ and $\vec d\perp\vec b$, $\vec d=t(32\hat i-\hat j-14\hat k)$ for some scalar $t$.

Now use $\vec c\cdot\vec d=15$ with $\vec c=2\hat i-\hat j+4\hat k$: …

(i) Cross product of $\vec a+\vec b$ and $\vec a-\vec b$, then normalize; (ii) expand the dot product using known magnitudes/dot product of $\vec a,\vec b$.

Given $\vec a = 3\hat i+2\hat j+2\hat k$, $\vec b = \hat i+2\hat j-2\hat k$.

(i) $\vec a+\vec b = 4\hat i+4\hat j+0\hat k$, $\vec a-\vec b = 2\hat i+0\hat j+4\hat k$.

$(\vec a+\vec b)\times(\vec a-\vec b) = \begin{vmatrix}\hat i&\hat j&\hat k\\4&4&0\\2&0&4\end{vmatrix} = \hat i(4\cdot4-0\cdot0) - \hat j(4\cdot4-0\cdot2) + \hat k(4\cdot0-4\cdot2) = 16\hat i-16\hat j-8\hat k.$

Magnitude $=\sqrt{16^2+16^2+8^2}=\sqrt{256+256+64}=\sqrt{576}=24$.

Unit vector $=\dfrac{16\hat i-16\hat j-8\hat k}{24} = \dfrac{2\hat i-2\hat j-\hat k}{3}$.

(ii) First find $|\vec a|^2 = 9+4+4=17$, $|\vec b|^2=1+4+4=9$, $\vec a\cdot\vec b = 3(1)+2(2)+2(-2)=3+4-4=3$.

$(3\vec a-5\vec b)\cdot(2\vec a+7\vec b) = 6\vec a\cdot\vec a + 21\vec a\cdot\vec b - 10\vec b\cdot\vec a - 35\vec b\cdot\vec b = 6|\vec a|^2 + 11(\vec a\cdot\vec b) - 35|\vec b|^2$

$= 6(17) + 11(3) - 35(9) = 102+33-315 = -180.$

OR: $A(1,-2,-8)$, $B(5,0,-2)$, $C(11,3,7)$.

…

A unit vector perpendicular to two vectors is their normalized cross product.

$\vec a=3\hat i+2\hat j+2\hat k$, $\vec b=\hat i+2\hat j-2\hat k$.

$\vec a+\vec b = 4\hat i+4\hat j+0\hat k$, $\quad \vec a-\vec b = 2\hat i+0\hat j+4\hat k$

$(\vec a+\vec b)\times(\vec a-\vec b) = \begin{vmatrix}\hat i&\hat j&\hat k\\4&4&0\\2&0&4\end{vmatrix}$

$=\hat i(4\cdot4-0\cdot0)-\hat j(4\cdot4-0\cdot2)+\hat k(4\cdot0-4\cdot2)$

$=16\hat i-16\hat j-8\hat k$

Magnitude $=\sqrt{16^2+16^2+8^2}=\sqrt{256+256+64}=\sqrt{576}=24$

Unit vector $=\dfrac{16\hat i-16\hat j-8\hat k}{24}=\dfrac13(2\hat i-2\hat j-\hat k)$ (or its negative).

OR: $\vec a=\hat i+4\hat j+2\hat k$, $\vec b=3\hat i-2\hat j+7\hat k$, $\vec c=2\hat i-\hat j+4\hat k$. Find $\vec d$ perpendicular to both $\vec a,\vec b$ with $\vec c\cdot\vec d=15$.

Since $\vec d\perp\vec a$ and $\vec d\perp\vec b$, $\vec d$ is parallel to $\vec a\times\vec b$, so $\vec d=\lambda(\vec a\times\vec b)$.

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