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Vector Algebra question

2023 · Shift 1 · Q29
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Vector Algebra question

2023 · Shift 1 · Q29

JEE AdvancedMathematicsVector AlgebraNumerical+4 / −1
Let PPP be the plane 3x+2y+3z=16\sqrt{3} x+2 y+3 z=163​x+2y+3z=16 and let S={αi^+βj^+γk^:α2+β2+γ2=1S=\left\{\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}: \alpha^2+\beta^2+\gamma^2=1\right.S={αi^+βj^​+γk^:α2+β2+γ2=1 and the distance of (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ) from the plane PPP is 72}\left.\frac{7}{2}\right\}27​}. Let u⃗,v⃗\vec{u}, \vec{v}u,v and w⃗\vec{w}w be three distinct vectors in SSS such that ∣u⃗−v⃗∣=∣v⃗−w⃗∣=∣w⃗−u⃗∣|\vec{u}-\vec{v}|=|\vec{v}-\vec{w}|=|\vec{w}-\vec{u}|∣u−v∣=∣v−w∣=∣w−u∣. Let VVV be the volume of the parallelepiped determined by vectors u⃗,v⃗\vec{u}, \vec{v}u,v and w⃗\vec{w}w. Then the value of 803V\frac{80}{\sqrt{3}} V3​80​V is :
Numerical answer
View written solutionFree

Correct answer: 45

  1. Interpret the set SSS

A vector r⃗=(α,β,γ)\vec{r}=(\alpha,\beta,\gamma)r=(α,β,γ) belongs to SSS if:

  • it is a unit vector: α2+β2+γ2=1\alpha^2+\beta^2+\gamma^2=1α2+β2+γ2=1
  • its distance from the plane 3x+2y+3z=16\sqrt{3}x+2y+3z=163​x+2y+3z=16 is 72\dfrac{7}{2}27​.

The distance of (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) from the plane is

∣3α+2β+3γ−16∣(3)2+22+32=∣3α+2β+3γ−16∣4.\frac{|\sqrt{3}\alpha+2\beta+3\gamma-16|}{\sqrt{(\sqrt{3})^2+2^2+3^2}} =\frac{|\sqrt{3}\alpha+2\beta+3\gamma-16|}{4}.(3​)2+22+32​∣3​α+2β+3γ−16∣​=4∣3​α+2β+3γ−16∣​.

Since this equals 72\dfrac{7}{2}27​,

∣3α+2β+3γ−16∣=14.|\sqrt{3}\alpha+2\beta+3\gamma-16|=14.∣3​α+2β+3γ−16∣=14.

So,

3α+2β+3γ=30or2.\sqrt{3}\alpha+2\beta+3\gamma=30 \quad \text{or} \quad 2.3​α+2β+3γ=30or2.

But for a unit vector (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ),

3α+2β+3γ≤3+4+9⋅α2+β2+γ2=4.\sqrt{3}\alpha+2\beta+3\gamma \le \sqrt{3+4+9}\cdot \sqrt{\alpha^2+\beta^2+\gamma^2}=4.3​α+2β+3γ≤3+4+9​⋅α2+β2+γ2​=4.

Hence 303030 is impossible. Therefore,

3α+2β+3γ=2.\sqrt{3}\alpha+2\beta+3\gamma=2.3​α+2β+3γ=2.

Thus SSS is the intersection of:

  • the unit sphere centered at origin,
  • the plane 3x+2y+3z=2.\sqrt{3}x+2y+3z=2.3​x+2y+3z=2.

So SSS is a circle.

  1. Find the center and radius of this circle

The plane has normal vector

n⃗=(3,2,3),∣n⃗∣=4.\vec{n}=(\sqrt{3},2,3), \qquad |\vec{n}|=4.n=(3​,2,3),∣n∣=4.

The plane is

n⃗⋅r⃗=2.\vec{n}\cdot \vec{r}=2.n⋅r=2.

Distance of this plane from origin is

2∣n⃗∣=24=12.\frac{2}{|\vec{n}|}=\frac{2}{4}=\frac12.∣n∣2​=42​=21​.

Therefore, the circle cut from the unit sphere has radius

r=1−(12)2=34=32.r=\sqrt{1-\left(\frac12\right)^2}=\sqrt{\frac34}=\frac{\sqrt{3}}{2}.r=1−(21​)2​=43​​=23​​.

Its center is the foot of perpendicular from origin to the plane:

c⃗=2∣n⃗∣2n⃗=216n⃗=18(3,2,3).\vec{c}=\frac{2}{|\vec{n}|^2}\vec{n}=\frac{2}{16}\vec{n}=\frac18(\sqrt{3},2,3).c=∣n∣22​n=162​n=81​(3​,2,3).

So

∣c⃗∣=12.|\vec{c}|=\frac12.∣c∣=21​.
  1. Use the condition ∣u⃗−v⃗∣=∣v⃗−w⃗∣=∣w⃗−u⃗∣|\vec{u}-\vec{v}|=|\vec{v}-\vec{w}|=|\vec{w}-\vec{u}|∣u−v∣=∣v−w∣=∣w−u∣

Since u⃗,v⃗,w⃗\vec{u},\vec{v},\vec{w}u,v,w are three distinct points on the circle SSS and all pairwise distances are equal, they form an equilateral triangle inscribed in that circle.

For an equilateral triangle with circumradius rrr, side length is

a=3 r.a=\sqrt{3}\,r.a=3​r.

Here

a=3⋅32=32.a=\sqrt{3}\cdot \frac{\sqrt{3}}{2}=\frac32.a=3​⋅23​​=23​.
  1. Area of triangle formed by u⃗,v⃗,w⃗\vec{u},\vec{v},\vec{w}u,v,w

The area of an equilateral triangle of side aaa is

Δ=34a2.\Delta=\frac{\sqrt{3}}{4}a^2.Δ=43​​a2.

So

Δ=34(32)2=34⋅94=9316.\Delta=\frac{\sqrt{3}}{4}\left(\frac32\right)^2 =\frac{\sqrt{3}}{4}\cdot \frac94 =\frac{9\sqrt{3}}{16}.Δ=43​​(23​)2=43​​⋅49​=1693​​.
  1. Relate tetrahedron volume and parallelepiped volume

The vectors u⃗,v⃗,w⃗\vec{u},\vec{v},\vec{w}u,v,w are position vectors of three points lying in the plane n⃗⋅r⃗=2\vec{n}\cdot \vec{r}=2n⋅r=2. So the perpendicular distance of this plane from the origin is 12\dfrac1221​.

Hence the volume of tetrahedron formed by O,u⃗,v⃗,w⃗O,\vec{u},\vec{v},\vec{w}O,u,v,w is

Vtetra=13×Δ×12=Δ6.V_{\text{tetra}}=\frac13 \times \Delta \times \frac12 =\frac{\Delta}{6}.Vtetra​=31​×Δ×21​=6Δ​.

But also,

Vtetra=16∣u⃗⋅(v⃗×w⃗)∣.V_{\text{tetra}}=\frac{1}{6}|\vec{u}\cdot(\vec{v}\times \vec{w})|.Vtetra​=61​∣u⋅(v×w)∣.

Therefore the parallelepiped volume is

V=∣u⃗⋅(v⃗×w⃗)∣=6Vtetra=Δ.V=|\vec{u}\cdot(\vec{v}\times \vec{w})|=6V_{\text{tetra}}=\Delta.V=∣u⋅(v×w)∣=6Vtetra​=Δ.

Thus,

V=9316.V=\frac{9\sqrt{3}}{16}.V=1693​​.
  1. Compute the required value
803V=803⋅9316=80⋅916=5⋅9=45.\frac{80}{\sqrt{3}}V=\frac{80}{\sqrt{3}}\cdot \frac{9\sqrt{3}}{16}=\frac{80\cdot 9}{16}=5\cdot 9=45.3​80​V=3​80​⋅1693​​=1680⋅9​=5⋅9=45.

Therefore, the required integer is

45.\boxed{45}.45​.
  1. Comparison with stored answer

Stored correct answer = 454545.

Our derived answer also is 454545, so they agree.

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