Let P be the plane 3x+2y+3z=16 and let S=αi^+βj^+γk^: α2+β2+γ2=1 and the distance of α, β, γ from the plane P is 72}. Let u→, v→ and w→ be three distinct vectors in S such that u→-v→=v→-w→=w→-u→. Let V be the volume of the parallelepiped determined by vectors u→, v→ and w→. Then the value of 803V is

JEE Advanced · Mathematics · 30. Vector Algebra

Let $P$ be the plane $\sqrt{3} x + 2 y + 3 z = 16$ and let
$S = \{α \hat{i} + β \hat{j} + γ \hat{k} : α^{2} + β^{2} + γ^{2} = 1$ and the distance of $(α, β, γ)$ from the plane $P$ is $\frac{7}{2}}$ .
Let $\vec{u}, \vec{v}$ and $\vec{w}$ be three distinct vectors in $S$ such that $|\vec{u} - \vec{v}| = |\vec{v} - \vec{w}| = |\vec{w} - \vec{u}|$ . Let $V$ be the volume of the parallelepiped determined by vectors $\vec{u}, \vec{v}$ and $\vec{w}$ . Then the value of $\frac{80}{\sqrt{3}} V$ is

A 15
B 35
C 45
D 72

Verified Solution

Answer & Solution

Correct Answer

(C) 45

Step-by-step Solution

Detailed explanation

Given,
Equation of plane,

P : \sqrt{3} x + 2 y + 3 z = 16

And

S = α \hat{i} + β \hat{j} + γ \hat{k} : α^{2} + β^{2} + γ^{2} = 1

which is equation of sphere,
Also distance,

d (α, β, γ)

from

P = \frac{7}{2}

And relation between distinct vector is given by,

|\vec{u} - \vec{v}| = |\vec{v} - \vec{w}| = |\vec{w} - \vec{u}|

Now

V

: volume of parallelepiped by vectors

\vec{u}, \vec{v}, \vec{w}

So, using formula of distance

d (α, β, γ)

from

P = \frac{7}{2}

we get,

\Rightarrow \frac{|\sqrt{3} α + 2 β + 3 γ - 16|}{\sqrt{3 + 4 + 9}} = \frac{7}{2}

\Rightarrow \frac{|\sqrt{3} α + 2 β + 3 γ - 16|}{4} = \frac{7}{2}

\Rightarrow |\sqrt{3} α + 2 β + 3 γ - 16| = 14 . . . . (i)

Also,

α^{2} + β^{2} + γ^{2} = 1 . . . . . . . (ii)

Now we know that,
Volume of parallelepiped by vector,

\vec{u}, \vec{v}, \vec{w}

is given by,

V = [\vec{u} \vec{v} \vec{w}]

= \vec{u} \cdot (v \times w) . . . . . . (iii)

|\vec{u}| = |\vec{v}| = |\vec{w}| = 1

(As they lie on sphere of unit radius)

. . . . . . (iv)

|\vec{u} - \vec{v}| = |\vec{v} - \vec{w}| = |\vec{w} - \vec{u}|

(Given)
Now squaring, we get,

\Rightarrow {|\vec{u} - \vec{v}|}^{2} = {|\vec{v} - \vec{w}|}^{2} = {|\vec{w} - \vec{u}|}^{2}

\Rightarrow \underset{(A)}{u^{2} + v^{2} - 2 \vec{u} \cdot \vec{v}} = \underset{(B)}{v^{2} + w^{2} - 2 \vec{v} \cdot \vec{w}} = \underset{(C)}{\underset{⏟}{w^{2} + u^{2} - 2 \vec{w} \cdot \vec{u}}}

Now from

(A)

and

(B)

we get,

\Rightarrow u^{2} + v^{2} - 2 \vec{u} \cdot \vec{v} = v^{2} + w^{2} - 2 \vec{v} \cdot \vec{w}

\Rightarrow u^{2} - w^{2} = 2 \vec{u} \cdot \vec{v} - 2 \vec{v} \cdot \vec{w}

[∵ |\vec{u}| = |\vec{w}| = 1 (Given)]

\Rightarrow \vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{w}

Hence, by using

(B)

and

(C)

also, we will get

\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{w} = \vec{w} \cdot \vec{u} = m (say) . . . . . . . (v)

\Rightarrow \vec{u}, \vec{v}, \vec{w}

are the vectors of an equilateral triangle (say

∆ A B C

)

d (O, P) = \frac{16}{\sqrt{3 + 4 + 9}}

= \frac{16}{4}

= 4

units

\vec{O A} = \vec{u}, \vec{O B} = \vec{v}, \vec{O C} = \vec{w}

|\vec{O A}| = |\vec{O B}| = |\vec{O C}| = 1

(Given)
In an equilateral triangle, circumcentre, orthrocentre and centroid coincide.
Let

D

be the circumcentre of

∆ A B C

, then

∠ A D B = 120^{\circ}

\Rightarrow \cos ∠ A D B = \frac{D A^{2} + D B^{2} - A B^{2}}{2 (D A) \cdot (D B)} . . . . . . (vi)

O E = O D + D E

= O D + A F

\Rightarrow 4 = O D + \frac{7}{2}

\Rightarrow O D = 4 - \frac{7}{2} = \frac{1}{2}

\Rightarrow D A = \sqrt{O A^{2} - O D^{2}}

= \sqrt{1 - \frac{1}{4}}

\Rightarrow D A = \frac{\sqrt{3}}{2}

\Rightarrow D A = D B = \frac{\sqrt{3}}{2} . . . . . (vii)

From

(vi)

and

(vii)

we get,

- \frac{1}{2} = \frac{\frac{3}{4} + \frac{3}{4} - A B^{2}}{2 \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}}

- \frac{1}{2} = \frac{\frac{3}{2} - A B^{2}}{\frac{3}{2}}

\Rightarrow - \frac{1}{2} \times \frac{3}{2} = \frac{3}{2} - A B^{2}

\Rightarrow A B^{2} = \frac{3}{2} + \frac{3}{4}

\Rightarrow A B^{2} = \frac{9}{4}

\Rightarrow A B = \frac{3}{2} = |\vec{u} - \vec{v}|

\Rightarrow A B^{2} = \frac{9}{4} = u^{2} + v^{2} - 2 \vec{u} \cdot \vec{v}

\Rightarrow \frac{9}{4} = 1 + 1 - 2 m

\Rightarrow 2 m = 2 - \frac{9}{4} = - \frac{1}{4}

\Rightarrow m = - \frac{1}{8} . . . . . . (viii)

Volume of parallelepiped,

V = |[\vec{u} \vec{v} \vec{w}]|

{|\vec{u} \vec{v} \vec{w}|}^{2} = |\begin{matrix} 1 & \vec{u} \cdot \vec{v} & \vec{u} \cdot \vec{w} \\ \vec{u} \cdot \vec{v} & 1 & \vec{v} \cdot \vec{w} \\ \vec{w} \cdot \vec{u} & \vec{w} \cdot \vec{v} & 1 \end{matrix}|

= |\begin{matrix} 1 & m & m \\ m & 1 & m \\ m & m & 1 \end{matrix}|

= 1 (1 - m^{2}) - m (m - m^{2}) + m (m^{2} - m)

= 1 - m^{2} - m^{2} + m^{3} + m^{3} - m^{2}

= 1 - 3 m^{2} + 2 m^{3}

{|\begin{array}{l} \vec{u} & \vec{v} & \vec{w} \end{array}|}^{2} = 2 m^{3} - 3 m^{2} + 1

= (m - 1) [2 m^{2} - m - 1]

= (m - 1) [2 m^{2} - 2 m + m - 1]

= (m - 1) (m - 1) (2 m + 1)

= {(m - 1)}^{2} (2 m + 1)

\Rightarrow |[\begin{array}{l} \vec{u} & \vec{v} & \vec{w} \end{array}]| = (m - 1) \sqrt{(2 m + 1)}

\Rightarrow V = |(- \frac{1}{8} - 1) \sqrt{2 \times - \frac{1}{8} + 1}|

\Rightarrow V = \frac{9}{8} \times \frac{\sqrt{3}}{2}

Hence,

\frac{80}{\sqrt{3}} V = \frac{80}{\sqrt{3}} \times \frac{9}{8} \times \frac{\sqrt{3}}{2}

= 45

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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