Here we have vectors a ⃗ = i ˆ + 3 j ˆ + k ˆ \vec{a}= \text{\^{i}}+3\text{\^{j}}+\text{\^{k}} a = i ˆ + 3 j ˆ + k ˆ , b ⃗ = 2 i ˆ + 4 j ˆ + 2 k ˆ \vec{b}=2\text{\^{i}}+4\text{\^{j}}+2\text{\^{k}} b = 2 i ˆ + 4 j ˆ + 2 k ˆ and c ⃗ = i ˆ + 2 j ˆ + 4 k ˆ \vec{c}=\text{\^{i}}+2\text{\^{j}}+4\text{\^{k}} c = i ˆ + 2 j ˆ + 4 k ˆ
So, we calculate d ⃗ = a ⃗ + b ⃗ + λ c ⃗ \vec{d}=\vec{a}+\vec{b}+\lambda\vec{c} d = a + b + λ c
d ⃗ = a ⃗ + b ⃗ + λ c ⃗ = i ˆ + 3 j ˆ + k ˆ + 2 i ˆ + 4 j ˆ + 2 k ˆ + λ i ˆ + 2 λ j ˆ + 4 λ k ˆ = ( 3 + λ ) i ˆ + ( 7 + 2 λ ) j ˆ + ( 3 + 4 λ ) k ˆ \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}} d = a + b + λ c = i ˆ + 3 j ˆ + k ˆ + 2 i ˆ + 4 j ˆ + 2 k ˆ + λ i ˆ + 2 λ j ˆ + 4 λ k ˆ = ( 3 + λ ) i ˆ + ( 7 + 2 λ ) j ˆ + ( 3 + 4 λ ) k ˆ
So, we have d ⃗ = ( 3 + λ ) i ˆ + ( 7 + 2 λ ) j ˆ + ( 3 + 4 λ ) k ˆ \vec{d}=(3+\lambda)\text{\^{i}}+(7+2\lambda)\text{\^{j}}+(3+4\lambda)\text{\^{k}} d = ( 3 + λ ) i ˆ + ( 7 + 2 λ ) j ˆ + ( 3 + 4 λ ) k ˆ
Now, we know that d ⃗ \vec{d} d parallel to y z yz yz plane. This means that d ⃗ \vec{d} d is ⊥ \perp ⊥ to x x x axis.
So, here we can equate d ⃗ ⋅ ( i ˆ ) = 0 \vec{d}\cdot(\text{\^{i}})=0 d ⋅ ( i ˆ ) = 0 [As the vector representation of x x x plane is i ˆ \text{\^{i}} i ˆ ]
So,
d ⃗ ⋅ i ˆ = 0 ⇒ [ ( ( 3 + λ ) i ˆ + ( 7 + 2 λ ) j ˆ + ( 3 + 4 λ ) k ˆ ] ⋅ [ i ˆ ] = 0 ⇒ 3 + λ = 0 ⇒ λ = − 3 \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 d ⋅ i ˆ = 0 ⇒ [(( 3 + λ ) i ˆ + ( 7 + 2 λ ) j ˆ + ( 3 + 4 λ ) k ˆ ] ⋅ [ i ˆ ] = 0 ⇒ 3 + λ = 0 ⇒ λ = − 3
So, value of λ \lambda λ such that it is parallel to y z yz yz plane is
λ = − 3 \fcolorbox{black}{aqua}{$\textcolor{black}{\lambda=-3}$} λ = − 3
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