Key Concepts and Formulas

• Cross Product Properties:
  • $\overrightarrow u \times \overrightarrow v$ is a vector perpendicular to both $\overrightarrow u$ and $\overrightarrow v$. This implies $(\overrightarrow u \times \overrightarrow v) \cdot \overrightarrow u = 0$ and $(\overrightarrow u \times \overrightarrow v) \cdot \overrightarrow v = 0$.
  • $|\overrightarrow u \times \overrightarrow v| = |\overrightarrow u| |\overrightarrow v| \sin \theta$, where $\theta$ is the angle between $\overrightarrow u$ and $\overrightarrow v$. If $\overrightarrow u \perp \overrightarrow v$, then $\theta = 90^\circ$, so $|\overrightarrow u \times \overrightarrow v| = |\overrightarrow u| |\overrightarrow v|$.
  • $\overrightarrow u \times \overrightarrow v = -(\overrightarrow v \times \overrightarrow u)$.
• Scalar Triple Product (STP): $[\overrightarrow u \ \overrightarrow v \ \overrightarrow w] = (\overrightarrow u \times \overrightarrow v) \cdot \overrightarrow w$. Cyclical permutation preserves the value: $[\overrightarrow u \ \overrightarrow v \ \overrightarrow w] = [\overrightarrow v \ \overrightarrow w \ \overrightarrow u] = [\overrightarrow w \ \overrightarrow u \ \overrightarrow v]$. If vectors are mutually orthogonal, $[\overrightarrow u \ \overrightarrow v \ \overrightarrow w] = \pm |\overrightarrow u| |\overrightarrow v| |\overrightarrow w|$.
• Vector Triple Product (VTP): $\overrightarrow u \times (\overrightarrow v \times \overrightarrow w) = (\overrightarrow u \cdot \overrightarrow w) \overrightarrow v - (\overrightarrow u \cdot \overrightarrow v) \overrightarrow w$.
• Magnitude Squared: For mutually orthogonal vectors $\overrightarrow x, \overrightarrow y, \overrightarrow z$, $|k_1 \overrightarrow x + k_2 \overrightarrow y + k_3 \overrightarrow z|^2 = k_1^2 |\overrightarrow x|^2 + k_2^2 |\overrightarrow y|^2 + k_3^2 |\overrightarrow z|^2$.
• Projection of $\overrightarrow u$ on $\overrightarrow v$: $\text{proj}_{\overrightarrow v} \overrightarrow u = \frac{\overrightarrow u \cdot \overrightarrow v}{|\overrightarrow v|}$.

Step-by-Step Solution

Step 1: Deduce mutual orthogonality and magnitudes of the vectors. We are given $\overrightarrow a \times \overrightarrow b = \overrightarrow c$ and $\overrightarrow b \times \overrightarrow c = \overrightarrow a$. From $\overrightarrow a \times \overrightarrow b = \overrightarrow c$, we know that $\overrightarrow c$ is perpendicular to both $\overrightarrow a$ and $\overrightarrow b$. Thus, $\overrightarrow c \cdot \overrightarrow a = 0$ and $\overrightarrow c \cdot \overrightarrow b = 0$. From $\overrightarrow b \times \overrightarrow c = \overrightarrow a$, we know that $\overrightarrow a$ is perpendicular to both $\overrightarrow b$ and $\overrightarrow c$. Thus, $\overrightarrow a \cdot \overrightarrow b = 0$ and $\overrightarrow a \cdot \overrightarrow c = 0$. Therefore, $\overrightarrow a, \overrightarrow b, \overrightarrow c$ are mutually orthogonal.

Since they are mutually orthogonal, for the cross product magnitudes, we have $|\overrightarrow x \times \overrightarrow y| = |\overrightarrow x| |\overrightarrow y|$. From $\overrightarrow a \times \overrightarrow b = \overrightarrow c$, we get $|\overrightarrow a| |\overrightarrow b| = |\overrightarrow c|$. From $\overrightarrow b \times \overrightarrow c = \overrightarrow a$, we get $|\overrightarrow b| |\overrightarrow c| = |\overrightarrow a|$. We are given $|\overrightarrow a| = 2$. Substituting $|\overrightarrow a| = 2$ into the equations:

1. $2 |\overrightarrow b| = |\overrightarrow c|$
2. $|\overrightarrow b| |\overrightarrow c| = 2$ Substitute (1) into (2): $|\overrightarrow b| (2|\overrightarrow b|) = 2 \implies 2|\overrightarrow b|^2 = 2 \implies |\overrightarrow b|^2 = 1 \implies |\overrightarrow b| = 1$. Substitute $|\overrightarrow b| = 1$ back into (1): $2(1) = |\overrightarrow c| \implies |\overrightarrow c| = 2$. So, the magnitudes are $|\overrightarrow a| = 2$, $|\overrightarrow b| = 1$, and $|\overrightarrow c| = 2$.

Step 2: Analyze Option (A). We need to check if $\overrightarrow a \times \left( {(\overrightarrow b + \overrightarrow c ) \times (\overrightarrow b \times \overrightarrow c )} \right) = \overrightarrow 0$. Using $\overrightarrow b \times \overrightarrow c = \overrightarrow a$, the expression becomes: $\overrightarrow a \times \left( {(\overrightarrow b + \overrightarrow c ) \times \overrightarrow a } \right)$ Using the distributive property of cross product: $= \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow a + \overrightarrow c \times \overrightarrow a } \right)$ We know $\overrightarrow b \times \overrightarrow a = -(\overrightarrow a \times \overrightarrow b) = -\overrightarrow c$. To find $\overrightarrow c \times \overrightarrow a$, we can use the VTP formula $\overrightarrow x \times (\overrightarrow y \times \overrightarrow z) = (\overrightarrow x \cdot \overrightarrow z) \overrightarrow y - (\overrightarrow x \cdot \overrightarrow y) \overrightarrow z$. Consider $\overrightarrow c \times (\overrightarrow b \times \overrightarrow c) = \overrightarrow c \times \overrightarrow a$. Here, $\overrightarrow x = \overrightarrow c, \overrightarrow y = \overrightarrow b, \overrightarrow z = \overrightarrow c$. $\overrightarrow c \times \overrightarrow a = (\overrightarrow c \cdot \overrightarrow c) \overrightarrow b - (\overrightarrow c \cdot \overrightarrow b) \overrightarrow c = |\overrightarrow c|^2 \overrightarrow b - 0 \cdot \overrightarrow c = |\overrightarrow c|^2 \overrightarrow b$. Since $|\overrightarrow c| = 2$, we have $\overrightarrow c \times \overrightarrow a = 2^2 \overrightarrow b = 4\overrightarrow b$. Substituting back into the expression: $= \overrightarrow a \times \left( {-\overrightarrow c + 4\overrightarrow b } \right)$ $= \overrightarrow a \times (-\overrightarrow c) + \overrightarrow a \times (4\overrightarrow b)$ $= -(\overrightarrow a \times \overrightarrow c) + 4(\overrightarrow a \times \overrightarrow b)$ We know $\overrightarrow a \times \overrightarrow c = -(\overrightarrow c \times \overrightarrow a) = -(4\overrightarrow b) = -4\overrightarrow b$. And $\overrightarrow a \times \overrightarrow b = \overrightarrow c$. So, the expression becomes: $= -(-4\overrightarrow b) + 4(\overrightarrow c) = 4\overrightarrow b + 4\overrightarrow c$. Since $\overrightarrow b$ and $\overrightarrow c$ are non-zero and orthogonal, $4\overrightarrow b + 4\overrightarrow c \neq \overrightarrow 0$. Therefore, Option (A) is not true.

Step 3: Analyze Option (B). The projection of $\overrightarrow a$ on $(\overrightarrow b \times \overrightarrow c)$ is given by $\frac{\overrightarrow a \cdot (\overrightarrow b \times \overrightarrow c)}{|\overrightarrow b \times \overrightarrow c|}$. We know $\overrightarrow b \times \overrightarrow c = \overrightarrow a$. So, the projection is $\frac{\overrightarrow a \cdot \overrightarrow a}{|\overrightarrow a|} = \frac{|\overrightarrow a|^2}{|\overrightarrow a|} = |\overrightarrow a|$. Since $|\overrightarrow a| = 2$, the projection is 2. Therefore, Option (B) is true.

Step 4: Analyze Option (C). We need to check if $[\overrightarrow a \ \overrightarrow b \ \overrightarrow c] + [\overrightarrow c \ \overrightarrow a \ \overrightarrow b] = 8$. Using the property of cyclic permutation of STP, $[\overrightarrow c \ \overrightarrow a \ \overrightarrow b] = [\overrightarrow a \ \overrightarrow b \ \overrightarrow c]$. So, the expression is $2 [\overrightarrow a \ \overrightarrow b \ \overrightarrow c]$. $[\overrightarrow a \ \overrightarrow b \ \overrightarrow c] = (\overrightarrow a \times \overrightarrow b) \cdot \overrightarrow c$. Given $\overrightarrow a \times \overrightarrow b = \overrightarrow c$, so $[\overrightarrow a \ \overrightarrow b \ \overrightarrow c] = \overrightarrow c \cdot \overrightarrow c = |\overrightarrow c|^2$. Since $|\overrightarrow c| = 2$, $|\overrightarrow c|^2 = 4$. Thus, $2 [\overrightarrow a \ \overrightarrow b \ \overrightarrow c] = 2 \times 4 = 8$. Therefore, Option (C) is true.

Step 5: Analyze Option (D). We need to check if $|3\overrightarrow a + \overrightarrow b - 2\overrightarrow c |^2 = 51$. Since $\overrightarrow a, \overrightarrow b, \overrightarrow c$ are mutually orthogonal, we can use the property $|k_1 \overrightarrow x + k_2 \overrightarrow y + k_3 \overrightarrow z|^2 = k_1^2 |\overrightarrow x|^2 + k_2^2 |\overrightarrow y|^2 + k_3^2 |\overrightarrow z|^2$. $|3\overrightarrow a + \overrightarrow b - 2\overrightarrow c |^2 = (3|\overrightarrow a|)^2 + (1|\overrightarrow b|)^2 + (-2|\overrightarrow c|)^2$ $= 9|\overrightarrow a|^2 + |\overrightarrow b|^2 + 4|\overrightarrow c|^2$. Substituting the magnitudes $|\overrightarrow a| = 2, |\overrightarrow b| = 1, |\overrightarrow c| = 2$: $= 9(2^2) + (1^2) + 4(2^2)$ $= 9(4) + 1 + 4(4)$ $= 36 + 1 + 16 = 53$. The value is 53, not 51. Therefore, Option (D) is not true.

Common Mistakes & Tips

• Confusing Orthogonality: Always confirm mutual orthogonality first, as it drastically simplifies magnitude calculations for sums of vectors.
• Vector Triple Product Errors: Be careful with the order of vectors and the dot products in the VTP formula.
• Sign Errors in Cross Products: Remember that $\overrightarrow u \times \overrightarrow v = -(\overrightarrow v \times \overrightarrow u)$.

Summary

The problem requires deducing the orthogonality and magnitudes of the given vectors from the cross product conditions. We found that $\overrightarrow a, \overrightarrow b, \overrightarrow c$ are mutually orthogonal with magnitudes $|\overrightarrow a|=2, |\overrightarrow b|=1, |\overrightarrow c|=2$. We then evaluated each option: Option (A) resulted in $4\overrightarrow b + 4\overrightarrow c$, which is not the zero vector. Option (B) correctly calculated the projection as 2. Option (C) correctly evaluated the sum of scalar triple products to 8. Option (D) calculated the squared magnitude to be 53, not 51. Since the question asks for the statement that is NOT true, and Option (A) is demonstrably false, it is the answer.

The final answer is $\boxed{A}$.