JEE Main 2024 — Mathematics Vectors & 3D Geometry

Given: $\vec{a}=\hat{i}+\hat{j}+k$a=i^+j^​+k, $\vec{b}=-\hat{i}-8\hat{j}+2\hat{k}$b=−i^−8j^​+2k^ and $\vec{c}=4\hat{i}+{c}_{2}\hat{j}+{c}_{3}k$c=4i^+c2​j^​+c3​k

Also, $\vec{b}\times \vec{a}=\vec{c}\times \vec{a}$b×a=c×a

$$\Rightarrow (\vec{b}-\vec{c})\times \vec{a}=0$$⇒(b−c)×a=0

$$\Rightarrow \vec{b}-\vec{c}=\lambda \vec{a}$$⇒b−c=λa

$$\Rightarrow \vec{b}=\vec{c}+\lambda \vec{a}$$⇒b=c+λa

$$\Rightarrow -\hat{i}-8\hat{j}+2k=(4\hat{i}+{c}_{2}\hat{j}+{c}_{3}k)+\lambda (\hat{i}+\hat{j}+k)$$⇒−i^−8j^​+2k=(4i^+c2​j^​+c3​k)+λ(i^+j^​+k)

$$\Rightarrow \lambda +4=-1,\lambda +{c}_{2}=-8,\lambda +{c}_{3}=2$$⇒λ+4=−1,λ+c2​=−8,λ+c3​=2

$$\Rightarrow \lambda =-5,{c}_{2}=-3,{c}_{3}=7$$⇒λ=−5,c2​=−3,c3​=7

$$\Rightarrow \vec{c}=4\hat{i}-3\hat{j}+7k$$⇒c=4i^−3j^​+7k

Now, finding angle we get,

$$\mathrm{cos}\theta =\frac{(3\hat{i}+4\hat{j}+\hat{k})\cdot (4\hat{i}-3\hat{j}+7\hat{k})}{|(3\hat{i}+4\hat{j}+\hat{k})|\cdot |(4\hat{i}-3\hat{j}+7\hat{k})|}$$cosθ=∣(3i^+4j^​+k^)∣⋅∣(4i^−3j^​+7k^)∣(3i^+4j^​+k^)⋅(4i^−3j^​+7k^)​

$$\Rightarrow \mathrm{cos}\theta =\frac{12-12+7}{\sqrt{26}\cdot \sqrt{74}}$$⇒cosθ=26​⋅74​12−12+7​

$$\Rightarrow \mathrm{cos}\theta =\frac{7}{\sqrt{26}\cdot \sqrt{74}}$$⇒cosθ=26​⋅74​7​

$$\Rightarrow \mathrm{cos}\theta =\frac{7}{2\sqrt{481}}$$⇒cosθ=2481​7​

$$\Rightarrow \mathrm{sec}\theta =\frac{2\sqrt{481}}{7}$$⇒secθ=72481​​

$$\Rightarrow {\mathrm{sec}}^{2}\theta =\frac{4\times 481}{49}$$⇒sec2θ=494×481​

$$\Rightarrow 1+{\mathrm{tan}}^{2}\theta =\frac{1924}{49}$$⇒1+tan2θ=491924​

$$\Rightarrow {\mathrm{tan}}^{2}\theta =\frac{1875}{49}$$⇒tan2θ=491875​

$$\Rightarrow {\mathrm{tan}}^{2}\theta \approx 38.26$$⇒tan2θ≈38.26

$$\Rightarrow [{\mathrm{tan}}^{2}\theta ]=[38.26]=38$$⇒[tan2θ]=[38.26]=38