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Mathematics · 3D Geometry

JEE Advanced 2019 — Paper 1 — Question 46

Three lines are given by r⃗=λi^,λ∈R,r⃗=μ(i^+j^),μ∈R\vec{r}=\lambda \hat{i}, \lambda \in R, \vec{r}=\mu(\hat{i}+\hat{j}), \mu \in R and r⃗=v(i^+j^+k^),v∈R\vec{r}=v(\hat{i}+\hat{j}+\hat{k}), v \in R. Let the

lines cut the plane x+y+z=1x+y+z=1 at the points A,BA, B and CC respectively. If the area of the triangle ABCA B C is

Δ\Delta then the value of (6Δ)2(6 \Delta)^{2} equals

Answer: 0.75

Numerical answer — enter this value.

Step-by-step solution

O is origin point C will be foot of perpendicular from O to plane

so C(13,13,13)\quad \mathrm{C}\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)

so, AB→=−12i^+12j^\quad \overrightarrow{\mathrm{AB}}=-\frac{1}{2} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}

AC→=−23i^+13j^+13k^\overrightarrow{\mathrm{AC}}=-\frac{2}{3} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{3} \hat{\mathrm{k}}

Δ=12∣AB→×AC→∣=12∣i^6+j^6+k^6∣=312\Delta=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\frac{1}{2}\left|\frac{\hat{\mathrm{i}}}{6}+\frac{\hat{\mathrm{j}}}{6}+\frac{\hat{\mathrm{k}}}{6}\right|=\frac{\sqrt{3}}{12}

(6Δ)2=34=0.75(6 \Delta)^{2}=\frac{3}{4}=0.75

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2019
Paper
Paper 1
Subject
Mathematics
Chapter
3D Geometry
Topic
Straight Lines in 3D Geometry

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