Let \overrightarrow a = 2\widehat i + \widehat j -2 \widehat k and \overrightarrow b = \widehat i + \widehat j . Let \overrightarrow c be a vector such that | \overrightarrow c - \overrightarrow a | = 3 , | ( \overrightarrow a \times \overrightarrow b ) \times \overrightarrow c | = 3 and the angle between \overrightarrow c and \overrightarrow a \times \overrightarrow b is 30^\circ . Then \overrightarrow a .\overrightarrow c is equal to :

Vector Algebra PYQ — JEE Main 2017 · Shift 0 · Q35 with Solution

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Correct answer: A

$\vec a = 2\hat i + \hat j - 2\hat k = (2,1,-2), \qquad \vec b = \hat i + \hat j = (1,1,0)$

We need to find $\vec a \cdot \vec c$ using the conditions:

$|\vec c - \vec a| = 3$ $|((\vec a \times \vec b) \times \vec c)| = 3$ Angle between $\vec c$ and $\vec a \times \vec b$ is $30^\circ$ .

\begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix}$$ $$= \hat i(1\cdot 0 - (-2)\cdot 1) - \hat j(2\cdot 0 - (-2)\cdot 1) + \hat k(2\cdot 1 - 1\cdot 1)$$ $$= 2\hat i - 2\hat j + \hat k$$ So, $$\vec d = \vec a \times \vec b = (2,-2,1)$$ and $$|\vec d| = \sqrt{2^2+(-2)^2+1^2} = \sqrt{9}=3$$ --- 3. **Use the triple cross-product magnitude condition** We know $$|\vec d \times \vec c| = 3$$ Also, angle between $\vec c$ and $\vec d$ is $30^\circ$. Hence $$|\vec d \times \vec c| = |\vec d|\,|\vec c|\sin 30^\circ$$ $$3 = 3\cdot |\vec c| \cdot \frac12$$ $$|\vec c| = 2$$ --- 4. **Use $|\vec c - \vec a| = 3$** Square both sides: $$|\vec c - \vec a|^2 = 9$$ Using $$|\vec c - \vec a|^2 = |\vec c|^2 + |\vec a|^2 - 2\vec a\cdot \vec c$$ First compute $|\vec a|^2$: $$|\vec a|^2 = 2^2+1^2+(-2)^2 = 9$$ and $|\vec c|^2 = 2^2=4$. So, $$9 = 4 + 9 - 2\vec a\cdot \vec c$$ $$9 = 13 - 2\vec a\cdot \vec c$$ $$2\vec a\cdot \vec c = 4$$ $$\vec a\cdot \vec c = 2$$ --- 5. **Compare with options** Thus, $$\boxed{\vec a\cdot \vec c = 2}$$ So the correct option is **A**. --- 6. **Verification with stored answer** Stored correct answer: **A** Our derived answer: **A** They agree.

View written solutionFree

Correct answer: A

$\vec a = 2\hat i + \hat j - 2\hat k = (2,1,-2), \qquad \vec b = \hat i + \hat j = (1,1,0)$

We need to find $\vec a \cdot \vec c$ using the conditions:

$|\vec c - \vec a| = 3$ $|((\vec a \times \vec b) \times \vec c)| = 3$ Angle between $\vec c$ and $\vec a \times \vec b$ is $30^\circ$ .

\begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix}$$ $$= \hat i(1\cdot 0 - (-2)\cdot 1) - \hat j(2\cdot 0 - (-2)\cdot 1) + \hat k(2\cdot 1 - 1\cdot 1)$$ $$= 2\hat i - 2\hat j + \hat k$$ So, $$\vec d = \vec a \times \vec b = (2,-2,1)$$ and $$|\vec d| = \sqrt{2^2+(-2)^2+1^2} = \sqrt{9}=3$$ --- 3. **Use the triple cross-product magnitude condition** We know $$|\vec d \times \vec c| = 3$$ Also, angle between $\vec c$ and $\vec d$ is $30^\circ$. Hence $$|\vec d \times \vec c| = |\vec d|\,|\vec c|\sin 30^\circ$$ $$3 = 3\cdot |\vec c| \cdot \frac12$$ $$|\vec c| = 2$$ --- 4. **Use $|\vec c - \vec a| = 3$** Square both sides: $$|\vec c - \vec a|^2 = 9$$ Using $$|\vec c - \vec a|^2 = |\vec c|^2 + |\vec a|^2 - 2\vec a\cdot \vec c$$ First compute $|\vec a|^2$: $$|\vec a|^2 = 2^2+1^2+(-2)^2 = 9$$ and $|\vec c|^2 = 2^2=4$. So, $$9 = 4 + 9 - 2\vec a\cdot \vec c$$ $$9 = 13 - 2\vec a\cdot \vec c$$ $$2\vec a\cdot \vec c = 4$$ $$\vec a\cdot \vec c = 2$$ --- 5. **Compare with options** Thus, $$\boxed{\vec a\cdot \vec c = 2}$$ So the correct option is **A**. --- 6. **Verification with stored answer** Stored correct answer: **A** Our derived answer: **A** They agree.