Let \overrightarrow a and \overrightarrow b be the vectors along the diagonals of a parallelogram having area 2\sqrt 2 . Let the angle between \overrightarrow a and \overrightarrow b be acute, |\overrightarrow a | = 1 , and |\overrightarrow a \,.\,\overrightarrow b | = |\overrightarrow a \times \overrightarrow b | . If \overrightarrow c = 2\sqrt 2 ( \overrightarrow a \times \overrightarrow b ) - 2\overrightarrow b , then an angle between \overrightarrow b and \overrightarrow c is :

Vector Algebra PYQ — JEE Main 2022 · 27 Jun · Shift 2 · Q32 with Solution

View written solutionFree

Correct answer: D

Use the diagonal-area relation for a parallelogram

If the diagonals of a parallelogram are represented by vectors $\vec a$ and $\vec b$ , then the area of the parallelogram is

$\frac12\,|\vec a\times \vec b|.$

Given area $=2\sqrt2$ , we get

\implies |\vec a\times \vec b|=4\sqrt2.$$ --- 2. **Use the condition** $|\vec a\cdot \vec b|=|\vec a\times \vec b|$ Since the angle between $\vec a$ and $\vec b$ is acute, $\vec a\cdot \vec b>0$. Hence $$\vec a\cdot \vec b=|\vec a\times \vec b|=4\sqrt2.$$ So, $$\vec a\cdot \vec b=4\sqrt2.$$ Also $|\vec a|=1$. --- 3. **Find** $|\vec b|$ Let the acute angle between $\vec a$ and $\vec b$ be $\theta$. Then $$\vec a\cdot \vec b=|\vec a||\vec b|\cos\theta=|\vec b|\cos\theta,$$ $$|\vec a\times \vec b|=|\vec a||\vec b|\sin\theta=|\vec b|\sin\theta.$$ Given these are equal and $\theta$ is acute, $$|\vec b|\cos\theta=|\vec b|\sin\theta \implies \cos\theta=\sin\theta \implies \theta=\frac\pi4.$$ Now $$|\vec a\times \vec b|=|\vec b|\sin\frac\pi4=|\vec b|\cdot \frac1{\sqrt2}=4\sqrt2.$$ Thus $$|\vec b|=8.$$ So far, $$\theta=\frac\pi4,\qquad |\vec b|=8,\qquad \vec a\cdot \vec b=4\sqrt2.$$ --- 4. **Interpret the expression for** $\vec c$ The given $$\vec c=2\sqrt2(\vec a\times \vec b)-2\vec b$$ must be treated in the usual vector-algebra sense where $\vec a\times\vec b$ is perpendicular to the plane of $\vec a,\vec b$. So $\vec c$ is a vector formed by adding a component perpendicular to $\vec b$ and a component along $-\vec b$. Let $$\vec u=2\sqrt2(\vec a\times \vec b), \qquad \vec v=-2\vec b.$$ Then $\vec u\perp \vec b$, hence $\vec u\perp \vec v$ as well. Now, $$|\vec u|=2\sqrt2\,|\vec a\times \vec b|=2\sqrt2\cdot 4\sqrt2=16,$$ and $$|\vec v|=2|\vec b|=16.$$ Thus $\vec c=\vec u+\vec v$ is the sum of two perpendicular vectors of equal magnitude $16$. Therefore $\vec c$ makes an angle of $45^\circ$ with $\vec v=-2\vec b$. So the angle between $\vec c$ and $-\vec b$ is $$\frac\pi4.$$ Hence the angle between $\vec c$ and $\vec b$ is $$\pi-\frac\pi4=\frac{3\pi}{4}.$$ --- 5. **Check options** - A: $\frac\pi4$ — angle with $-\vec b$, not with $\vec b$ - B: $-\frac\pi4$ — not an angle between vectors - C: $\frac{5\pi}{6}$ — incorrect - D: $\frac{3\pi}{4}$ — correct So the required angle is $$\boxed{\frac{3\pi}{4}}.$$

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View written solutionFree

Correct answer: D

Use the diagonal-area relation for a parallelogram

If the diagonals of a parallelogram are represented by vectors $\vec a$ and $\vec b$ , then the area of the parallelogram is

$\frac12\,|\vec a\times \vec b|.$

Given area $=2\sqrt2$ , we get

\implies |\vec a\times \vec b|=4\sqrt2.$$ --- 2. **Use the condition** $|\vec a\cdot \vec b|=|\vec a\times \vec b|$ Since the angle between $\vec a$ and $\vec b$ is acute, $\vec a\cdot \vec b>0$. Hence $$\vec a\cdot \vec b=|\vec a\times \vec b|=4\sqrt2.$$ So, $$\vec a\cdot \vec b=4\sqrt2.$$ Also $|\vec a|=1$. --- 3. **Find** $|\vec b|$ Let the acute angle between $\vec a$ and $\vec b$ be $\theta$. Then $$\vec a\cdot \vec b=|\vec a||\vec b|\cos\theta=|\vec b|\cos\theta,$$ $$|\vec a\times \vec b|=|\vec a||\vec b|\sin\theta=|\vec b|\sin\theta.$$ Given these are equal and $\theta$ is acute, $$|\vec b|\cos\theta=|\vec b|\sin\theta \implies \cos\theta=\sin\theta \implies \theta=\frac\pi4.$$ Now $$|\vec a\times \vec b|=|\vec b|\sin\frac\pi4=|\vec b|\cdot \frac1{\sqrt2}=4\sqrt2.$$ Thus $$|\vec b|=8.$$ So far, $$\theta=\frac\pi4,\qquad |\vec b|=8,\qquad \vec a\cdot \vec b=4\sqrt2.$$ --- 4. **Interpret the expression for** $\vec c$ The given $$\vec c=2\sqrt2(\vec a\times \vec b)-2\vec b$$ must be treated in the usual vector-algebra sense where $\vec a\times\vec b$ is perpendicular to the plane of $\vec a,\vec b$. So $\vec c$ is a vector formed by adding a component perpendicular to $\vec b$ and a component along $-\vec b$. Let $$\vec u=2\sqrt2(\vec a\times \vec b), \qquad \vec v=-2\vec b.$$ Then $\vec u\perp \vec b$, hence $\vec u\perp \vec v$ as well. Now, $$|\vec u|=2\sqrt2\,|\vec a\times \vec b|=2\sqrt2\cdot 4\sqrt2=16,$$ and $$|\vec v|=2|\vec b|=16.$$ Thus $\vec c=\vec u+\vec v$ is the sum of two perpendicular vectors of equal magnitude $16$. Therefore $\vec c$ makes an angle of $45^\circ$ with $\vec v=-2\vec b$. So the angle between $\vec c$ and $-\vec b$ is $$\frac\pi4.$$ Hence the angle between $\vec c$ and $\vec b$ is $$\pi-\frac\pi4=\frac{3\pi}{4}.$$ --- 5. **Check options** - A: $\frac\pi4$ — angle with $-\vec b$, not with $\vec b$ - B: $-\frac\pi4$ — not an angle between vectors - C: $\frac{5\pi}{6}$ — incorrect - D: $\frac{3\pi}{4}$ — correct So the required angle is $$\boxed{\frac{3\pi}{4}}.$$

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