Let O be the origin and OA⃗ = 2î + 2ĵ + k̂, OB⃗ = î - 2ĵ + 2k̂ and OC⃗ = 1/2(OB⃗ - λOA⃗) for some λ > 0. If |OB⃗ × OA⃗| = 9/2, then which of the following statements is (are) TRUE?

Let O be the origin and OA⃗ = 2î + 2ĵ + k̂, OB⃗ = î - 2ĵ + 2k̂ and OC⃗ = 1/2(OB⃗ - λOA⃗) for some λ > 0. If |OB⃗ × OA⃗| = 9/2, then which of the following statements is (are) TRUE? Given: OA⃗ = 2î + 2ĵ + k̂, OB⃗ = î - 2ĵ + 2k̂ OC⃗ = 1/2(OB⃗ - λOA⃗), λ > 0 |OB⃗ × OA⃗| = 9/2 To find: which statements are TRUE? (1) proj₍OA) OC⃗ = -3/2 (2) [OAB] = 9/2 (3) [ABC] = 9/2 (4) diagonals of OA,OC parallelogram = π/3 OC⃗ = 1/2(OB⃗ - λOA⃗), λ > 0 OA⃗·OB⃗ = (2)(1) + (2)(-2) + (1)(2) OA⃗·OB⃗ = 2 - 4 + 2 = 0 ⇒ OA⃗⊥OB⃗ |OA⃗| = √(4+4+1) = 3, |OB⃗| = √(1+4+4) = 3 OB⃗×OA⃗ = î ĵ k̂ 1 -2 2 2 2 1 OB⃗×OA⃗ = (-6, 3, 6) |OB⃗×OA⃗| = √(36+9+36) = 9 [OAB] = 1/2|OA⃗×OB⃗| = 9/2 OB⃗×OC⃗ = 1/2(OB⃗×OB⃗ - λ(OB⃗×OA⃗)) = -λ/2(OB⃗×OA⃗) |OB⃗×OC⃗| = λ/2· 9 = 9λ/2 = 9/2 ⇒ λ = 1 λ = 1: OC⃗ = 1/2(OB⃗ - OA⃗) = 1/2(-1, -4, 1) OC⃗ = (-1/2, -2, 1/2) Plan: 1. proj₍OA) OC⃗ 2. [OAB]; 3. [ABC] = 1/2|AB⃗×AC⃗| 4. angle between diagonals of the OA,OC parallelogram proj₍OA)OC⃗ = (OC⃗·OA⃗)/|OA⃗| OC⃗·OA⃗ = -1/2(2) + (-2)(2) + 1/2(1) = -9/2 proj = (-9/2)/3 = -3/2 checkmark [OAB] = 1/2|OA⃗×OB⃗| = 1/2(3)(3) = 9/2 checkmark AB⃗ = B - A = (1-2, -2-2, 2-1) = (-1, -4, 1) AC⃗ = C - A = (-1/2-2, -2-2, 1/2-1) = (-5/2, -4, -1/2) AB⃗×AC⃗ = î ĵ k̂ -1 -4 1 -5/2 -4 -1/2 = î(2+4) - ĵ(1/2+5/2) + k̂(4-10) AB⃗×AC⃗ = (6, -3, -6) |AB⃗×AC⃗| = √(36+9+36) = 9 [ABC] = 1/2|AB⃗×AC⃗| = 9/2 checkmark d₁⃗ = OA⃗ + OC⃗ = (3/2, 0, 3/2) d₂⃗ = OA⃗ - OC⃗ = (5/2, 4, 1/2) d₁⃗·d₂⃗ = (3/2)(5/2) + 0 + (3/2)(1/2) = 9/2 |d₁⃗| = 3√2/2, |d₂⃗| = 3√(10)/2 cosθ = (9/2)/(3√2/2·3√(10)/2) = 1/√5 ⇒ θ ≈ 63.4^∘ ≠ π/3 u2717 boxedTrue: (1), (2), (3)

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

Answers: A. Projection of $\overrightarrow{OC}$ on $\overrightarrow{OA}$ is $-\frac{3}{2}$ · B. Area of the triangle $OAB$ is $\frac{9}{2}$ · C. Area of the triangle $ABC$ is $\frac{9}{2}$

Vector Algebra Let

O

be the origin and

\overrightarrow{OA} = 2\hat{i} + 2\hat{j} + \hat{k}

\overrightarrow{OB} = \hat{i} - 2\hat{j} + 2\hat{k}

and

\overrightarrow{OC} = \frac{1}{2}(\overrightarrow{OB} - \lambda\overrightarrow{OA})

for some

\lambda > 0

. If

|\overrightarrow{OB} \times \overrightarrow{OA}| = \frac{9}{2}

, then which of the following statements is (are) TRUE?

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Written solution

$\text{Given: } \overrightarrow{OA} = 2\hat{i} + 2\hat{j} + \hat{k},\quad \overrightarrow{OB} = \hat{i} - 2\hat{j} + 2\hat{k}$ $\overrightarrow{OC} = \dfrac{1}{2}\big(\overrightarrow{OB} - \lambda\overrightarrow{OA}\big),\quad \lambda > 0$ $|\overrightarrow{OB} \times \overrightarrow{OA}| = \dfrac{9}{2}$ $\text{To find: which statements are TRUE?}$ $(1)\ \text{proj}_{OA}\,\overrightarrow{OC} = -\dfrac{3}{2}\qquad (2)\ [OAB] = \dfrac{9}{2}$ $(3)\ [ABC] = \dfrac{9}{2}\qquad (4)\ \text{diagonals of } OA,OC \text{ parallelogram} = \dfrac{\pi}{3}$ $\overrightarrow{OC} = \dfrac{1}{2}(\overrightarrow{OB} - \lambda\overrightarrow{OA}),\quad \lambda > 0$ $\overrightarrow{OA}\cdot\overrightarrow{OB} = (2)(1) + (2)(-2) + (1)(2)$ $\overrightarrow{OA}\cdot\overrightarrow{OB} = 2 - 4 + 2 = 0 \Rightarrow \overrightarrow{OA}\perp\overrightarrow{OB}$ $|\overrightarrow{OA}| = \sqrt{4+4+1} = 3,\qquad |\overrightarrow{OB}| = \sqrt{1+4+4} = 3$ $\overrightarrow{OB}\times\overrightarrow{OA} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 2 & 2 & 1 \end{vmatrix}$ $\overrightarrow{OB}\times\overrightarrow{OA} = (-6,\ 3,\ 6)$ $|\overrightarrow{OB}\times\overrightarrow{OA}| = \sqrt{36+9+36} = 9$ $[OAB] = \dfrac{1}{2}|\overrightarrow{OA}\times\overrightarrow{OB}| = \dfrac{9}{2}$ $\overrightarrow{OB}\times\overrightarrow{OC} = \dfrac{1}{2}\big(\overrightarrow{OB}\times\overrightarrow{OB} - \lambda(\overrightarrow{OB}\times\overrightarrow{OA})\big) = -\dfrac{\lambda}{2}(\overrightarrow{OB}\times\overrightarrow{OA})$ $|\overrightarrow{OB}\times\overrightarrow{OC}| = \dfrac{\lambda}{2}\cdot 9 = \dfrac{9\lambda}{2} = \dfrac{9}{2} \Rightarrow \lambda = 1$ $\lambda = 1:\quad \overrightarrow{OC} = \dfrac{1}{2}(\overrightarrow{OB} - \overrightarrow{OA}) = \dfrac{1}{2}(-1,\ -4,\ 1)$ $\overrightarrow{OC} = \left(-\dfrac{1}{2},\ -2,\ \dfrac{1}{2}\right)$ $\text{Plan: } 1.\ \text{proj}_{OA}\,\overrightarrow{OC}$ $2.\ [OAB];\qquad 3.\ [ABC] = \dfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|$ $4.\ \text{angle between diagonals of the } OA,OC \text{ parallelogram}$ $\text{proj}_{OA}\overrightarrow{OC} = \dfrac{\overrightarrow{OC}\cdot\overrightarrow{OA}}{|\overrightarrow{OA}|}$ $\overrightarrow{OC}\cdot\overrightarrow{OA} = -\dfrac{1}{2}(2) + (-2)(2) + \dfrac{1}{2}(1) = -\dfrac{9}{2}$ $\text{proj} = \dfrac{-9/2}{3} = -\dfrac{3}{2}\ \checkmark$ $[OAB] = \dfrac{1}{2}|\overrightarrow{OA}\times\overrightarrow{OB}| = \dfrac{1}{2}(3)(3) = \dfrac{9}{2}\ \checkmark$ $\overrightarrow{AB} = B - A = (1-2,\ -2-2,\ 2-1) = (-1,\ -4,\ 1)$ $\overrightarrow{AC} = C - A = \left(-\dfrac{1}{2}-2,\ -2-2,\ \dfrac{1}{2}-1\right) = \left(-\dfrac{5}{2},\ -4,\ -\dfrac{1}{2}\right)$ $\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -4 & 1 \\ -\frac{5}{2} & -4 & -\frac{1}{2} \end{vmatrix}$ $= \hat{i}(2+4) - \hat{j}\left(\dfrac{1}{2}+\dfrac{5}{2}\right) + \hat{k}(4-10)$ $\overrightarrow{AB}\times\overrightarrow{AC} = (6,\ -3,\ -6)$ $|\overrightarrow{AB}\times\overrightarrow{AC}| = \sqrt{36+9+36} = 9$ $[ABC] = \dfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}| = \dfrac{9}{2}\ \checkmark$ $\vec{d_1} = \overrightarrow{OA} + \overrightarrow{OC} = \left(\dfrac{3}{2},\ 0,\ \dfrac{3}{2}\right)$ $\vec{d_2} = \overrightarrow{OA} - \overrightarrow{OC} = \left(\dfrac{5}{2},\ 4,\ \dfrac{1}{2}\right)$ $\vec{d_1}\cdot\vec{d_2} = \left(\dfrac{3}{2}\right)\left(\dfrac{5}{2}\right) + 0 + \left(\dfrac{3}{2}\right)\left(\dfrac{1}{2}\right) = \dfrac{9}{2}$ $|\vec{d_1}| = \dfrac{3\sqrt{2}}{2},\qquad |\vec{d_2}| = \dfrac{3\sqrt{10}}{2}$ $\cos\theta = \dfrac{9/2}{\frac{3\sqrt2}{2}\cdot\frac{3\sqrt{10}}{2}} = \dfrac{1}{\sqrt{5}} \Rightarrow \theta \approx 63.4^\circ \neq \dfrac{\pi}{3}\ \text{✗}$ $\boxed{\text{True: } (1),\ (2),\ (3)}$

Written solution

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