OABC is a tetrahedron in which O is the origin and position vector of points A, B, C are i^+2 j^+3 k^, 2 i^ +α j^ + k^ and i^ + 3 j^ +2 k^ respectively. A value of ‘α ’ for which shortest distance between OA and BC is 32

Unit vector n^ perpendicular to OA→and CB→is in n^ =3α-7 i^-4 j^+5-α k^3α-72+16+5-α2 ∴32=BA n^⇒α=97or 3OA→=i^+2 j^+3 k^,OB→=2 i^+α j^+k^ and OC→=i^+3 j^+2 k^⇒BA→=OA→−OB→=−i^+(2−α) j^+2 k^,CB→=OB→−OC→=i^+(α−3) j^−k^, and Unit vector ( n^ ) perpendicular to OA→ and CB→ is n^=CB→×OA→|CB→×OA→|=3α-7 i^-4 j^+5-α k^3α-72+16+5-α2 As, CB→×OA→=i^j^k^1α−3−1123=(3α−7)i^−4j^+(5−α)k^Shortest distance=d=|BA→⋅n^|⇒32=(7−3a)−4(2−α)+2(5−α)(3α−7)2+(16)+(5−α)2⇒32=|−α+9|10α2−52α+90squaring and simplifying7α2−30α+27=0⇒(7α−9)(α−3)=0⇒α=97,3

OABC is a tetrahedron in which O is the origin and position vector of points A, B, C are i^+2 j^+3 k^, 2 i^ +α j^ + k^ and i^ + 3 j^ +2 k^ respectively. A value of ‘α ’ for which shortest distance between OA and BC is 32

Unit vector n^ perpendicular to OA→and CB→is in n^ =3α-7 i^-4 j^+5-α k^3α-72+16+5-α2 ∴32=BA n^⇒α=97or 3OA→=i^+2 j^+3 k^,OB→=2 i^+α j^+k^ and OC→=i^+3 j^+2 k^⇒BA→=OA→−OB→=−i^+(2−α) j^+2 k^,CB→=OB→−OC→=i^+(α−3) j^−k^, and Unit vector ( n^ ) perpendicular to OA→ and CB→ is n^=CB→×OA→|CB→×OA→|=3α-7 i^-4 j^+5-α k^3α-72+16+5-α2 As, CB→×OA→=i^j^k^1α−3−1123=(3α−7)i^−4j^+(5−α)k^Shortest distance=d=|BA→⋅n^|⇒32=(7−3a)−4(2−α)+2(5−α)(3α−7)2+(16)+(5−α)2⇒32=|−α+9|10α2−52α+90squaring and simplifying7α2−30α+27=0⇒(7α−9)(α−3)=0⇒α=97,3

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OABC is a tetrahedron in which O is the origin and position vector of points A, B, C are $\hat{i} + 2 \hat{j} + 3 \hat{k}, 2 \hat{i} + α \hat{j} + \hat{k} a n d \hat{i} + 3 \hat{j} + 2 \hat{k}$ respectively. A value of ‘ $α$ ’ for which shortest distance between OA and BC is $\sqrt{\frac{3}{2}}$

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$\frac{3}{7}$

$- 3$

$\frac{9}{7}$

$\frac{- 3}{7}$

answer is B.

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Detailed Solution

Unit vector $(\hat{n})$ perpendicular to $\vec{OA} and \vec{CB} is in \hat{n} = \frac{(3 α - 7) \hat{i} - 4 \hat{j} + (5 - α) \hat{k}}{\sqrt{{(3 α - 7)}^{2} + 16 + {(5 - α)}^{2}}} ∴ \sqrt{\frac{3}{2}} = |BA \hat{n}| \Rightarrow α = \frac{9}{7} or 3$

$\begin{array}{l} \vec{O A} = \hat{i} + 2 \hat{j} + 3 \hat{k}, \vec{O B} = 2 \hat{i} + α \hat{j} + \hat{k} and \vec{O C} = \hat{i} + 3 \hat{j} + 2 \hat{k} \\ \Rightarrow \vec{B A} = \vec{O A} - \vec{O B} = - \hat{i} + (2 - α) \hat{j} + 2 \hat{k}, \\ \vec{C B} = \vec{O B} - \vec{O C} = \hat{i} + (α - 3) \hat{j} - \hat{k} \\ , and \\ Unit vector (\hat{n}) perpendicular to \vec{O A} and \vec{C B} is \hat{n} = \frac{\vec{CB} \times \vec{OA}}{| \vec{CB} \times \vec{OA} |} \\ = \frac{(3 α - 7) \hat{i} - 4 \hat{j} + (5 - α) \hat{k}}{\sqrt{{(3 α - 7)}^{2} + 16 + {(5 - α)}^{2}}} \\ As, \vec{C B} \times \vec{O A} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & α - 3 & - 1 \\ 1 & 2 & 3 \end{array}| = (3 α - 7) \hat{i} - 4 \hat{j} + (5 - α) \hat{k} \\ \begin{array}{l} S h o r t e s t d i s \tan c e = d = | \vec{B A} \cdot \hat{n} | \\ \Rightarrow \sqrt{\frac{3}{2}} = |\frac{(7 - 3 a) - 4 (2 - α) + 2 (5 - α)}{\sqrt{(3 α - 7)^{2} + (16) + (5 - α)^{2}}}| \\ \Rightarrow \sqrt{\frac{3}{2}} = \frac{| - α + 9 |}{\sqrt{10 α^{2} - 52 α + 90}} \\ s q u a r i n g a n d s i m p l i f y i n g \\ \begin{aligned} 7 α^{2} - 30 α + 27 = 0 \\ \Rightarrow (7 α - 9) (α - 3) = 0 \\ \Rightarrow α = \frac{9}{7}, 3 \end{aligned} \end{array} \end{array}$