Given vec u=6vec i+vec j+vec k , vec v=3vec j-vec k and vec…

Given $\vec u=6\vec i+\vec j+\vec k$, $\vec v=3\vec j-\vec k$ and $\vec w=-2\vec i+3\vec j+5\vec k$, determine $\lambda$ so that $\vec u+\lambda\vec v$ and $\vec w$ are perpendicular.

Explanation of Method

Find λ so that u+λv is perpendicular to w.

Step 1: The vector $\vec u+\lambda\vec v$

\[ \vec u+\lambda\vec v=(6,1,1)+\lambda(0,3,-1)=(6,\ 1+3\lambda,\ 1-\lambda). \]

Step 2: Perpendicularity condition

\[ (\vec u+\lambda\vec v)\cdot\vec w=6(-2)+3(1+3\lambda)+5(1-\lambda)=-4+4\lambda=0. \]

\[\lambda=1\]

Step 3: Check

\[ \vec u+\vec v=(6,4,0),\qquad (6,4,0)\cdot(-2,3,5)=-12+12+0=0.\ \checkmark \]

Both vectors are non-zero, $|\vec u+\vec v|=2\sqrt{13}$ and $|\vec w|=\sqrt{38}$, so the angle between them is exactly $90^\circ$.

$O$ $\vec u$ $\vec v$ $\vec w$ $\vec u+\vec v$

$\vec u+\lambda\vec v=(6,4,0)\ \ (\lambda=1)$
$\vec w=(-2,3,5)$

For $\lambda=1$ the vector $\vec u+\vec v$ lies in the $xOy$ plane (dotted) and is perpendicular to $\vec w$ (right-angle mark at $O$).

Result

\[\boxed{\,\lambda=1\,}\]

Remark

Alternative. $\vec u\cdot\vec w=-4$ and $\vec v\cdot\vec w=4$. The condition $\vec u\cdot\vec w+\lambda\,\vec v\cdot\vec w=0$ gives $\lambda=-\dfrac{\vec u\cdot\vec w}{\vec v\cdot\vec w}=1$ directly.