Question

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

Answer 1

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$A=ai+bj+ck$ :
$\left|A\right|=\sqrt{a^2+b^2+c^2}$