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QJEEM23Ma7cfb: JEE Main 2023 Mathematics - Class 12 JEEM23Ma7cfb Solution: Let a line pass through the origin and be perpendicular to the…

Let a line ll pass through the origin and be perpendicular to the lines

l1:r⃗=(ı^−11ȷ^−7k^)+λ(i^+2ȷ^+3k^),λ∈Rl_{1}: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in \mathbb{R} and

l2:r⃗=(−ı^+k^)+μ(2ı^+2ȷ^+k^),μ∈Rl_{2}: \vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k}}), \mu \in \mathbb{R} .

If P\mathrm{P} is the point of intersection of ll and l1l_{1} , and Q(∝,β,γ)\mathrm{Q}(\propto, \beta, \gamma) is the foot of perpendicular from P on l2l_{2} , then 9(α+β+γ)9(\alpha+\beta+\gamma) is equal to _____________.

Step-by-Step Verified Solution

We have,

l1:r⃗=(ı^−11ȷ^−7k^)+λ(i^+2ȷ^+3k^),λ∈Rl_{1}: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in \mathbb{R} and

l2:r⃗=(−ı^+k^)+μ(2ı^+2ȷ^+k^),μ∈Rl_{2}: \vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k}}), \mu \in \mathbb{R} .

Let direction ratio of line ll be a,ba, b and cc Equation of line ll

r⃗amp;=(0i^+0j^+0k^)+δ(ai^+bj^+ck^)amp;=δ(ai^+bj^+ck^)\begin{aligned}\vec{r} & =(0 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+0 \hat{\mathbf{k}})+\delta(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}) \\\\& =\delta(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}})\end{aligned}

As, line ll is perpendicular to l1l_1 and l2l_2 ,

ai^+bj^+ck^=∣i^amp;j^amp;k^1amp;2amp;32amp;2amp;1∣=−4i^+5j^−2k^a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\1 & 2 & 3 \\2 & 2 & 1\end{array}\right|=-4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}

∴\therefore Equation of line l:r⃗=δ(−4i^+5j^−2k^)l: \vec{r}=\delta(-4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})

As, PP is the intersecting point of ll and l1l_1

−4δ=1+λ,5δ=−11+2λ,−2δ=−7+3λ-4 \delta=1+\lambda, 5 \delta=-11+2 \lambda,-2 \delta=-7+3 \lambda

After solving the above three equation, we get

δ=−1 and λ=3\delta=-1 \text { and } \lambda=3

∴\therefore Co-ordinate of point PP is (4,−5,2)(4,-5,2) .

QQ is a point on line l2l_2

Let co-ordinate of QQ be (−1+2μ,2μ,1+μ)(-1+2 \mu, 2 \mu, 1+\mu)

PQ→=(−5+2μ)i^+(2μ+5)j^+(μ−1)k^PQ→⋅(2i^+2j^+k^)=0[∵PQ→⊥l2]2(−5+2μ)+2(2μ+5)+μ−1=09μ−1=0⇒μ=19\begin{gathered}\overrightarrow {PQ}=(-5+2 \mu) \hat{\mathbf{i}}+(2 \mu+5) \hat{\mathbf{j}}+(\mu-1) \hat{\mathbf{k}} \\\\\overrightarrow {PQ} \cdot(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})=0 \left[\because \overrightarrow {PQ} \perp l_2\right]\\\\2(-5+2 \mu)+2(2 \mu+5)+\mu-1=0 \\\\9 \mu-1=0 \Rightarrow \mu=\frac{1}{9}\end{gathered}

amp;∴α=−1+29=−79,β=2×19=29,r=1+19=109amp; Hence, 9(α+β+γ)=9(−79+29+109)=5\begin{aligned}& \therefore \alpha=-1+\frac{2}{9}=\frac{-7}{9}, \beta=2 \times \frac{1}{9}=\frac{2}{9}, r=1+\frac{1}{9}=\frac{10}{9} \\\\& \text { Hence, } 9(\alpha+\beta+\gamma)=9\left(-\frac{7}{9}+\frac{2}{9}+\frac{10}{9}\right)=5\end{aligned}

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