Answer to Question #159686 in Analytic Geometry for Muhammed

Question

7. If a = i + 3j j k, b = 2i + 4j j 2k and c = =i + 2j + 4k, find a number λ such that

d = a + b + λc is parallel to the yz-plane.

Answer 1

Here we have vectors $\vec{a}= \text{\^{i}}+3\text{\^{j}}+\text{\^{k}}$ , $\vec{b}=2\text{\^{i}}+4\text{\^{j}}+2\text{\^{k}}$ and $\vec{c}=\text{\^{i}}+2\text{\^{j}}+4\text{\^{k}}$

So, we calculate $\vec{d}=\vec{a}+\vec{b}+\lambda\vec{c}$

$$\vec{d}=\vec{a}+\vec{b}+\lambda\vec{c}\\ =\text{\^{i}}+3\text{\^{j}}+\text{\^{k}}+2\text{\^{i}}+4\text{\^{j}}+2\text{\^{k}}+\lambda\text{\^{i}}+2\lambda\text{\^{j}}+4\lambda\text{\^{k}}\\ =(3+\lambda)\text{\^{i}}+(7+2\lambda)\text{\^{j}}+(3+4\lambda)\text{\^{k}}$$

So, we have $\vec{d}=(3+\lambda)\text{\^{i}}+(7+2\lambda)\text{\^{j}}+(3+4\lambda)\text{\^{k}}$

Now, we know that $\vec{d}$ parallel to $yz$ plane. This means that $\vec{d}$ is $\perp$ to $x$ axis.

So, here we can equate $\vec{d}\cdot(\text{\^{i}})=0$ [As the vector representation of $x$ plane is $\text{\^{i}}$ ]

So,

$$\vec{d}\cdot\text{\^{i}}=0\\ \Rightarrow [((3+\lambda)\text{\^{i}}+(7+ 2\lambda)\text{\^{j}}+(3+4\lambda)\text{\^{k}}]\cdot[\text{\^{i}}]=0\\ \Rightarrow 3+\lambda=0\\ \Rightarrow \lambda=-3$$

So, value of $\lambda$ such that it is parallel to $yz$ plane is

$$\fcolorbox{black}{aqua}{$\textcolor{black}{\lambda=-3}$}$$