# Find $\lambda$ so that the vectors $\overrightarrow{2i}+\overrightarrow{\lambda\;j}+\overrightarrow{k}$ and $\overrightarrow{i}- \overrightarrow{2j} +\overrightarrow{k}$ are perpendicular to each other.

## 1 Answer

Toolbox:
• If $\overrightarrow a = a_1\overrightarrow i + a_2 \overrightarrow j+a_3 \overrightarrow k,\: \: \overrightarrow b = b_1 \overrightarrow i+b_2\overrightarrow j + b_3 \overrightarrow k$ then $\overrightarrow a.\overrightarrow b = a_1b_1+a_2b_2+a_3b_3$
• If $\overrightarrow a \perp \overrightarrow b$ then $\overrightarrow a.\overrightarrow b=0$ and for nonzero vectors if $\overrightarrow a.\overrightarrow b=0 \Rightarrow \overrightarrow a \perp \overrightarrow b.$
Step 1
Let $\overrightarrow a = 2\overrightarrow i+ \lambda \overrightarrow j+\overrightarrow k$ and $\overrightarrow b=\overrightarrow i-2\overrightarrow j+\overrightarrow k$
$\overrightarrow a.\overrightarrow b = (2)(1)+( \lambda)(-2)+(1)(1)=2-2\lambda+1=3-2\lambda$
Step 2
If $\overrightarrow a \perp \overrightarrow b$ then $\overrightarrow a.\overrightarrow b=0. \: \: \therefore 4-2\lambda = 0 \: \: or \: 2\lambda=3$
$\therefore \lambda = \large\frac{3}{2}$
answered May 31, 2013

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