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

Projection of \overrightarrow OC on \overrightarrow OA is - 3 \over 2, Area of the triangle OAB is 9 \over 2, Area of the triangle ABC is 9 \over 2

Projection of \overrightarrow OC on \overrightarrow OA is - 3 \over 2

Area of the triangle OAB is 9 \over 2

Area of the triangle ABC is 9 \over 2

The acute angle between the diagonals of the parallelogram with adjacent sides \overrightarrow OA and \overrightarrow OC is \pi \over 3

Vector Algebra PYQ — JEE Advanced 2021 · Shift 2 · Q24 with Solution

View written solutionFree

Correct answer: A, B, C

Given vectors

$\vec{OA}=\langle 2,2,1\rangle,\qquad \vec{OB}=\langle 1,-2,2\rangle$

and

$\vec{OC}=\frac12(\vec{OB}-\lambda \vec{OA}), \qquad \lambda>0.$

So,

$\vec{OC}=\frac12\big(\langle 1,-2,2\rangle-\lambda\langle 2,2,1\rangle\big)$

-2-2\lambda, 2-\lambda\rangle =\left\langle \frac{1-2\lambda}{2},-(1+\lambda),\frac{2-\lambda}{2}\right\rangle.$$ --- 2. **Use the condition** $|\vec{OB}\times \vec{OC}|=\dfrac92$ Since $$\vec{OC}=\frac12(\vec{OB}-\lambda\vec{OA}),$$ we get $$\vec{OB}\times \vec{OC}=\frac12\left(\vec{OB}\times \vec{OB}-\lambda(\vec{OB}\times \vec{OA})\right).$$ But $\vec{OB}\times \vec{OB}=\vec{0}$, so $$\vec{OB}\times \vec{OC}=-\frac\lambda2(\vec{OB}\times \vec{OA}).$$ Hence, $$|\vec{OB}\times \vec{OC}|=\frac\lambda2|\vec{OB}\times \vec{OA}|.$$ Now compute $\vec{OA}\times \vec{OB}$: $$\vec{OA}\times \vec{OB}= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2&2&1\\ 1&-2&2 \end{vmatrix} =\hat i(4+2)-\hat j(4-1)+\hat k(-4-2)$$ $$=6\hat i-3\hat j-6\hat k.$$ Therefore, $$|\vec{OA}\times \vec{OB}|=\sqrt{6^2+(-3)^2+(-6)^2}=\sqrt{81}=9.$$ So also $|\vec{OB}\times \vec{OA}|=9$. Thus, $$\frac\lambda2\cdot 9=\frac92$$ $$\Rightarrow \lambda=1.$$ --- 3. **Find** $\vec{OC}$ Putting $\lambda=1$: $$\vec{OC}=\frac12(\vec{OB}-\vec{OA})$$ $$=\frac12\big(\langle 1,-2,2\rangle-\langle 2,2,1\rangle\big) =\frac12\langle -1,-4,1\rangle =\left\langle -\frac12,-2,\frac12\right\rangle.$$ --- 4. **Check option A** Projection of $\vec{OC}$ on $\vec{OA}$ means scalar projection: $$\operatorname{proj}_{\vec{OA}}(\vec{OC})=\frac{\vec{OC}\cdot \vec{OA}}{|\vec{OA}|}.$$ First, $$\vec{OC}\cdot \vec{OA}=\left(-\frac12\right)(2)+(-2)(2)+\left(\frac12\right)(1)=-1-4+\frac12=-\frac92.$$ Also, $$|\vec{OA}|=\sqrt{2^2+2^2+1^2}=\sqrt9=3.$$ Hence, $$\frac{\vec{OC}\cdot \vec{OA}}{|\vec{OA}|}=\frac{-9/2}{3}=-\frac32.$$ So **A is true**. --- 5. **Check option B** Area of triangle $OAB$ is $$\frac12|\vec{OA}\times \vec{OB}|=\frac12\cdot 9=\frac92.$$ So **B is true**. --- 6. **Check option C** Area of triangle $ABC$ is $$\frac12|\overrightarrow{AB}\times \overrightarrow{AC}|.$$ Now, $$\overrightarrow{AB}=\vec{OB}-\vec{OA}, \qquad \overrightarrow{AC}=\vec{OC}-\vec{OA}.$$ Using $\vec{OC}=\dfrac12(\vec{OB}-\vec{OA})$, $$\overrightarrow{AB}=\vec{OB}-\vec{OA}=2\vec{OC}.$$ Also, $$\overrightarrow{AC}=\vec{OC}-\vec{OA}.$$ Let us compute directly: $$\overrightarrow{AB}=\langle -1,-4,1\rangle,$$ $$\overrightarrow{AC}=\left\langle -\frac12,-2,\frac12\right\rangle-\langle 2,2,1\rangle =\left\langle -\frac52,-4,-\frac12\right\rangle.$$ Then $$\overrightarrow{AB}\times \overrightarrow{AC}= \begin{vmatrix} \hat i&\hat j&\hat k\\ -1&-4&1\\ -5/2&-4&-1/2 \end{vmatrix}$$ $$=\hat i\left(2-(-4)\right)-\hat j\left(\frac12-\left(-\frac52\right)\right)+\hat k(4-10)$$ $$=6\hat i-3\hat j-6\hat k.$$ So, $$|\overrightarrow{AB}\times \overrightarrow{AC}|=9.$$ Therefore area of triangle $ABC$ is $$\frac12\cdot 9=\frac92.$$ So **C is true**. --- 7. **Check option D** For the parallelogram with adjacent sides $\vec{OA}=\vec a$ and $\vec{OC}=\vec c$, the diagonals are $$\vec d_1=\vec a+\vec c, \qquad \vec d_2=\vec a-\vec c.$$ Here, $$\vec a=\langle 2,2,1\rangle, \qquad \vec c=\left\langle -\frac12,-2,\frac12\right\rangle.$$ So, $$\vec d_1=\left\langle \frac32,0,\frac32\right\rangle, \qquad \vec d_2=\left\langle \frac52,4,\frac12\right\rangle.$$ Now, $$\vec d_1\cdot \vec d_2=\frac32\cdot\frac52+0\cdot 4+\frac32\cdot\frac12 =\frac{15}{4}+\frac34=\frac{18}{4}=\frac92.$$ Also, $$|\vec d_1|=\sqrt{\left(\frac32\right)^2+\left(\frac32\right)^2}=\frac{3}{\sqrt2},$$ $$|\vec d_2|=\sqrt{\left(\frac52\right)^2+4^2+\left(\frac12\right)^2} =\sqrt{\frac{25}{4}+16+\frac14} =\sqrt{\frac{90}{4}}=\frac{3\sqrt{10}}{2}.$$ Hence, $$\cos\theta=\frac{\vec d_1\cdot \vec d_2}{|\vec d_1||\vec d_2|} =\frac{9/2}{(3/\sqrt2)(3\sqrt{10}/2)} =\frac{9/2}{9\sqrt5/2}=\frac1{\sqrt5}.$$ Thus, $$\theta=\cos^{-1}\left(\frac1{\sqrt5}\right)\neq \frac\pi3.$$ So **D is false**. --- 8. **Final conclusion** The true statements are: $$\boxed{A,\ B,\ C}$$

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View written solutionFree

Correct answer: A, B, C

Given vectors

$\vec{OA}=\langle 2,2,1\rangle,\qquad \vec{OB}=\langle 1,-2,2\rangle$

and

$\vec{OC}=\frac12(\vec{OB}-\lambda \vec{OA}), \qquad \lambda>0.$

So,

$\vec{OC}=\frac12\big(\langle 1,-2,2\rangle-\lambda\langle 2,2,1\rangle\big)$

-2-2\lambda, 2-\lambda\rangle =\left\langle \frac{1-2\lambda}{2},-(1+\lambda),\frac{2-\lambda}{2}\right\rangle.$$ --- 2. **Use the condition** $|\vec{OB}\times \vec{OC}|=\dfrac92$ Since $$\vec{OC}=\frac12(\vec{OB}-\lambda\vec{OA}),$$ we get $$\vec{OB}\times \vec{OC}=\frac12\left(\vec{OB}\times \vec{OB}-\lambda(\vec{OB}\times \vec{OA})\right).$$ But $\vec{OB}\times \vec{OB}=\vec{0}$, so $$\vec{OB}\times \vec{OC}=-\frac\lambda2(\vec{OB}\times \vec{OA}).$$ Hence, $$|\vec{OB}\times \vec{OC}|=\frac\lambda2|\vec{OB}\times \vec{OA}|.$$ Now compute $\vec{OA}\times \vec{OB}$: $$\vec{OA}\times \vec{OB}= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2&2&1\\ 1&-2&2 \end{vmatrix} =\hat i(4+2)-\hat j(4-1)+\hat k(-4-2)$$ $$=6\hat i-3\hat j-6\hat k.$$ Therefore, $$|\vec{OA}\times \vec{OB}|=\sqrt{6^2+(-3)^2+(-6)^2}=\sqrt{81}=9.$$ So also $|\vec{OB}\times \vec{OA}|=9$. Thus, $$\frac\lambda2\cdot 9=\frac92$$ $$\Rightarrow \lambda=1.$$ --- 3. **Find** $\vec{OC}$ Putting $\lambda=1$: $$\vec{OC}=\frac12(\vec{OB}-\vec{OA})$$ $$=\frac12\big(\langle 1,-2,2\rangle-\langle 2,2,1\rangle\big) =\frac12\langle -1,-4,1\rangle =\left\langle -\frac12,-2,\frac12\right\rangle.$$ --- 4. **Check option A** Projection of $\vec{OC}$ on $\vec{OA}$ means scalar projection: $$\operatorname{proj}_{\vec{OA}}(\vec{OC})=\frac{\vec{OC}\cdot \vec{OA}}{|\vec{OA}|}.$$ First, $$\vec{OC}\cdot \vec{OA}=\left(-\frac12\right)(2)+(-2)(2)+\left(\frac12\right)(1)=-1-4+\frac12=-\frac92.$$ Also, $$|\vec{OA}|=\sqrt{2^2+2^2+1^2}=\sqrt9=3.$$ Hence, $$\frac{\vec{OC}\cdot \vec{OA}}{|\vec{OA}|}=\frac{-9/2}{3}=-\frac32.$$ So **A is true**. --- 5. **Check option B** Area of triangle $OAB$ is $$\frac12|\vec{OA}\times \vec{OB}|=\frac12\cdot 9=\frac92.$$ So **B is true**. --- 6. **Check option C** Area of triangle $ABC$ is $$\frac12|\overrightarrow{AB}\times \overrightarrow{AC}|.$$ Now, $$\overrightarrow{AB}=\vec{OB}-\vec{OA}, \qquad \overrightarrow{AC}=\vec{OC}-\vec{OA}.$$ Using $\vec{OC}=\dfrac12(\vec{OB}-\vec{OA})$, $$\overrightarrow{AB}=\vec{OB}-\vec{OA}=2\vec{OC}.$$ Also, $$\overrightarrow{AC}=\vec{OC}-\vec{OA}.$$ Let us compute directly: $$\overrightarrow{AB}=\langle -1,-4,1\rangle,$$ $$\overrightarrow{AC}=\left\langle -\frac12,-2,\frac12\right\rangle-\langle 2,2,1\rangle =\left\langle -\frac52,-4,-\frac12\right\rangle.$$ Then $$\overrightarrow{AB}\times \overrightarrow{AC}= \begin{vmatrix} \hat i&\hat j&\hat k\\ -1&-4&1\\ -5/2&-4&-1/2 \end{vmatrix}$$ $$=\hat i\left(2-(-4)\right)-\hat j\left(\frac12-\left(-\frac52\right)\right)+\hat k(4-10)$$ $$=6\hat i-3\hat j-6\hat k.$$ So, $$|\overrightarrow{AB}\times \overrightarrow{AC}|=9.$$ Therefore area of triangle $ABC$ is $$\frac12\cdot 9=\frac92.$$ So **C is true**. --- 7. **Check option D** For the parallelogram with adjacent sides $\vec{OA}=\vec a$ and $\vec{OC}=\vec c$, the diagonals are $$\vec d_1=\vec a+\vec c, \qquad \vec d_2=\vec a-\vec c.$$ Here, $$\vec a=\langle 2,2,1\rangle, \qquad \vec c=\left\langle -\frac12,-2,\frac12\right\rangle.$$ So, $$\vec d_1=\left\langle \frac32,0,\frac32\right\rangle, \qquad \vec d_2=\left\langle \frac52,4,\frac12\right\rangle.$$ Now, $$\vec d_1\cdot \vec d_2=\frac32\cdot\frac52+0\cdot 4+\frac32\cdot\frac12 =\frac{15}{4}+\frac34=\frac{18}{4}=\frac92.$$ Also, $$|\vec d_1|=\sqrt{\left(\frac32\right)^2+\left(\frac32\right)^2}=\frac{3}{\sqrt2},$$ $$|\vec d_2|=\sqrt{\left(\frac52\right)^2+4^2+\left(\frac12\right)^2} =\sqrt{\frac{25}{4}+16+\frac14} =\sqrt{\frac{90}{4}}=\frac{3\sqrt{10}}{2}.$$ Hence, $$\cos\theta=\frac{\vec d_1\cdot \vec d_2}{|\vec d_1||\vec d_2|} =\frac{9/2}{(3/\sqrt2)(3\sqrt{10}/2)} =\frac{9/2}{9\sqrt5/2}=\frac1{\sqrt5}.$$ Thus, $$\theta=\cos^{-1}\left(\frac1{\sqrt5}\right)\neq \frac\pi3.$$ So **D is false**. --- 8. **Final conclusion** The true statements are: $$\boxed{A,\ B,\ C}$$

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