Let O be the origin and OA→=2i^+2j^+k^, OB→=i^-2j^+2k^ and OC→=12OB→-λOA→ for some λ>0. If OB→×OC→=92, then which of the following statements is (are) TRUE?

(A) Projection of OC→ and OA→ is -32

(B) Area of the triangle OAB is 92

(C) Area of the triangle ABC is 92

(D) The acute angle between the diagonals of the parallelogram with adjacent sides OA→ and OC→ is π3

JEE Advanced · Mathematics · 30. Vector Algebra

Let $O$ be the origin and $\vec{O A} = 2 \hat{i} + 2 \hat{j} + \hat{k}, \vec{O B} = \hat{i} - 2 \hat{j} + 2 \hat{k}$ and $\vec{O C} = \frac{1}{2} (\vec{O B} - λ \vec{O A})$ for some $λ > 0 .$ If $|\vec{O B} \times \vec{O C}| = \frac{9}{2},$ then which of the following statements is (are) TRUE?

A Projection of $\vec{O C}$ and $\vec{O A}$ is $- \frac{3}{2}$
B Area of the triangle $O A B$ is $\frac{9}{2}$
C Area of the triangle $A B C$ is $\frac{9}{2}$
D The acute angle between the diagonals of the parallelogram with adjacent sides $\vec{O A}$ and $\vec{O C}$ is $\frac{π}{3}$

Verified Solution

Answer & Solution

Correct Answer

(A) Projection of $\vec{O C}$ and $\vec{O A}$ is $- \frac{3}{2}$

Step-by-step Solution

Detailed explanation

\vec{O A} = 2 \hat{i} + 2 \hat{j} + \hat{k}

\vec{O B} = \hat{i} - 2 \hat{j} + 2 \hat{k}

\vec{O C} = \frac{1}{2} (\vec{O B} - λ \vec{O A}); λ > 0

|\vec{O B} \times \vec{O C}| = \frac{9}{2}

\Rightarrow |\vec{O B} \times \frac{1}{2} (\vec{O B} - λ \vec{O A})| = \frac{9}{2}

\Rightarrow \frac{1}{2} |\vec{O B} \times (\vec{O B} - λ \vec{O A})| = \frac{9}{2}

\Rightarrow \frac{1}{2} |λ (\vec{O A} \times \vec{O B})| = \frac{9}{2}

\Rightarrow |\frac{λ}{2} (\vec{O A} \times \vec{O B})| = \frac{9}{2} . . . (1)

now,

\vec{O A} \times \vec{O B} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 1 & - 2 & 2 \end{matrix}|

= 6 \hat{i} - 3 \hat{j} - 6 \hat{k}

\Rightarrow |\vec{O A} \times \vec{O B}| = \sqrt{36 + 9 + 36} = 9 . . . (2)

Put in (1)

|\frac{λ}{2} \times 9| = \frac{9}{2}

\Rightarrow |λ| = 1

\Rightarrow λ = \pm 1 (∵ λ > 0)

\Rightarrow λ = 1

\Rightarrow \vec{O C} = \frac{1}{2} (\vec{O B} - \vec{O A})

\vec{O C} = \frac{1}{2} (- \hat{i} - 4 \hat{j} + \hat{k})

(A)
Projection of

\vec{O C}

\vec{O A}

= \frac{\vec{O C} \cdot \vec{O A}}{|\vec{O A}|}

= \frac{1}{2} \frac{(- \hat{i} - 4 \hat{j} + \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} + \hat{k})}{\sqrt{4 + 4 + 1}}

= \frac{1}{2} \frac{(- 2 - 8 + 1)}{3}

= \frac{- 9}{6} = - \frac{3}{2}

(B) Area of

Δ O A B = \frac{1}{2} |\vec{O A} \times \vec{O B}|

= \frac{9}{2}

sq unit
(C)
Area of

A B E = \frac{1}{2} | \vec{A B} \times \vec{A C} |

\vec{A B} = - \hat{i} - 4 \hat{j} + \hat{k}

\vec{A C} = - \frac{5}{2} \hat{i} - 4 \hat{j} - \frac{\hat{k}}{2}

\vec{A B} \times \vec{A C} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 1 & - 4 & 1 \\ - \frac{5}{2} & - 4 & - \frac{1}{2} \end{matrix}|

= 6 \hat{i} - 3 \hat{j} - 6 \hat{k}

| \vec{A B} \times \vec{A C} | = \sqrt{36 + 9 + 36} = 9

Area of

∆ A B C = \frac{1}{2} |\vec{A B} \times \vec{A C}|

= \frac{9}{2} sq . unit

(D)

Apply angle between diagonals

\vec{O D}

\vec{C A}

\cos θ = \frac{(\vec{a} + \vec{c}) \cdot (\vec{a} - \vec{c})}{|\vec{a} + \vec{c}| |\vec{a} - \vec{c}|}

\vec{a} = 2 \hat{i} + 2 \hat{j} + \hat{k}

\vec{c} = \frac{1}{2} (- \hat{i} + 4 \hat{j} + \hat{k})

\vec{a} + \vec{c} = \frac{3}{2} \hat{i} + \frac{3}{2} \hat{k}

|\vec{a} + \vec{c}| = \sqrt{\frac{9}{4} + \frac{9}{4}} = \frac{3 \sqrt{2}}{2}

\vec{a} - \vec{c} = \frac{5}{2} \hat{i} + 4 \hat{j} + \frac{1}{2} \hat{k}

|\vec{a} - \vec{c}| = \sqrt{\frac{25}{4} + 16 + \frac{1}{4}} = \sqrt{\frac{90}{4}} = \frac{3 \sqrt{10}}{2}

\cos θ = \frac{\frac{15}{4} + \frac{3}{4}}{(\frac{3 \sqrt{2}}{2}) (\frac{3 \sqrt{10}}{2})}

= \frac{\frac{9}{2}}{\frac{9}{4} \times \sqrt{20}} = \frac{1}{\sqrt{5}} \neq \frac{π}{3}

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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