For what value of $\lambda $, the vector $i - \lambda + 2k$ and $8i + 6j - k$ are at right angles?

Hint: According to the question the angle between two vectors is ${90^ \circ }$. Therefore, apply the Dot product formula for the angle between two Vectors.
$\overrightarrow a \cdot \overrightarrow b = \left| a \right|\left| b \right|\cos \theta $
Since $\theta = {90^ \circ }$ and $\cos {90^ \circ } = 0$
$\therefore \overrightarrow a \cdot \overrightarrow b = \left| a \right|\left| b \right|\cos {90^ \circ }$
$\therefore \overrightarrow a \cdot \overrightarrow b = 0$
By solving the equation $\overrightarrow a \cdot \overrightarrow b = 0$ we get the value of $\lambda $.

Complete step-by-step answer:
Consider the two vectors $i - \lambda + 2k$ and $8i + 6j - k$. They are at right angles, so $\theta = {90^ \circ }$.
If the two vectors are assumed as $\overrightarrow a $ and $\overrightarrow b $ then the dot product denoted as $\overrightarrow a \cdot \overrightarrow b $ . Suppose these two vectors are separated by angle θ.
The dot product of two product is given as
$\overrightarrow a \cdot \overrightarrow b = \left| a \right|\left| b \right|\cos \theta $
Substitute $\theta = {90^ \circ }$ and $\cos {90^ \circ } = 0$.
$\therefore \overrightarrow a \cdot \overrightarrow b = \left| a \right|\left| b \right|\cos {90^ \circ }$
$\therefore \overrightarrow a \cdot \overrightarrow b = 0$
Suppose $\overrightarrow a = i - \lambda + 2k$ and $\overrightarrow b = 8i + 6j - k$ then evaluate $\overrightarrow a \cdot \overrightarrow b = 0$.
$ \Rightarrow (i - \lambda + 2k) \cdot (8i + 6j - k) = 0$
Since $i \cdot i = 1$, $j \cdot j = 1$ and $k \cdot k = 1$ we have,
$ \Rightarrow 1 \times 8 - \lambda \times 6 + 2 \times ( - 1) = 0$
$ \Rightarrow 8 - 6\lambda - 2 = 0$
$ \Rightarrow 6 - 6\lambda = 0$
$ \Rightarrow 6\lambda = 6$
$ \Rightarrow \lambda = 1$

Final Answer: The vector $i - \lambda + 2k$ and $8i + 6j - k$ are at right angles for $\lambda = 1$.

Note:
Remember the difference between dot product and cross product.
The dot product of two vectors $A$ and $B$ is represented as: $A \cdot B = \left| A \right|\left| B \right|\cos \theta $, where $\left| A \right|$ and $\left| B \right|$ are the magnitude of the vectors.
The cross product of two vectors $A$ and $B$ is represented as: $A \times B = \left| A \right|\left| B \right|\sin \theta $, where$\left| A \right|$ and $\left| B \right|$ are the magnitude of the vectors.
Magnitude of the vector$A = ai + bj + ck$ :
$\left| A \right| = \sqrt {{a^2} + {b^2} + {c^2}} $