Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

Polygon law is applicable in both but the equation given in the reason is not useful in explaining the assertion.

Since perpendicular vectors must have a dot product of zero, we can represent the problem through two separate equations. Expanding $(\vec{a}+3\vec{b}) \cdot (7\vec{a}-5\vec{b}) = 0$ leads to the equation $7|\vec{a}|^2 + 16\vec{a}\cdot\vec{b} - 15|\vec{b}|^2 = 0$. Similarly, expanding the second condition $(\vec{a}-4\vec{b}) \cdot (7\vec{a}-2\vec{b}) = 0$ gives us $7|\vec{a}|^2 - 30\vec{a}\cdot\vec{b} + 8|\vec{b}|^2 = 0$.

Subtracting the second equation from the first allows us to eliminate the $|\vec{a}|^2$ term, resulting in $46\vec{a}\cdot\vec{b} - 23|\vec{b}|^2 = 0$, which simplifies to the relationship $\vec{a}\cdot\vec{b} = \frac{1}{2}|\vec{b}|^2$. By substituting this result back into either of our earlier equations, we find that $7|\vec{a}|^2 - 7|\vec{b}|^2 = 0$, confirming that $|\vec{a}| = |\vec{b}|$.

With these components established, we use the definition of the dot product $\cos \theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$. Substituting our findings, we get $\cos \theta = \frac{0.5|\vec{b}|^2}{|\vec{b}||\vec{b}|}$, which simplifies to $0.5$. This confirms that the angle between the two vectors is 60 degrees.

When equating the magnitudes of two vector expressions, the most effective approach is to square both sides to eliminate the square roots and utilize the properties of the dot product. By expanding both $|2a + 3b|^2$ and $|3a + b|^2$ using the identity $|u+v|^2 = |u|^2 + |v|^2 + 2(u \cdot v)$, we generate a relationship involving the squared magnitudes and the dot product of the vectors. Specifically, the expansion results in the equation:

$4|a|^2 + 9|b|^2 + 12(a \cdot b) = 9|a|^2 + |b|^2 + 6(a \cdot b)$

Rearranging these terms allows us to combine like factors, simplifying the expression to $5|a|^2 - 6(a \cdot b) - 8|b|^2 = 0$. Since $\frac{1}{8}a$ is a unit vector, we know $|a| = 8$. Furthermore, the dot product $a \cdot b$ is defined by $|a||b| \cos(60^{\circ})$, which simplifies to $8|b| \cdot \frac{1}{2}$, or $4|b|$. Substituting $|a| = 8$ and the result for the dot product into our simplified equation creates a quadratic expression in terms of $|b|$:

$5(64) - 6(4|b|) - 8|b|^2 = 0$

Dividing the entire equation by $-8$ transforms it into $|b|^2 + 3|b| - 40 = 0$, which factors into $(|b|+8)(|b|-5) = 0$. Given that the magnitude of a vector must be positive, we discard the negative root and find that the value of $|b|$ must be $5$.

The length of the projection of vector $\vec{b}$ onto vector $\vec{a} \times \vec{c}$ is defined by the expression $l = \frac{|\vec{b} \cdot (\vec{a} \times \vec{c})|}{|\vec{a} \times \vec{c}|}$. Using the scalar triple product property, the numerator can be rewritten as $(\vec{a} \times \vec{b}) \cdot \vec{c}$. Since the problem states that $\vec{a} \times \vec{b} = \vec{c}$, the numerator simplifies to $\vec{c} \cdot \vec{c}$, or $|\vec{c}|^2$. Given $\vec{c} = \hat{j} - \hat{k}$, its squared magnitude is $0^2 + 1^2 + (-1)^2 = 2$.

To determine the denominator, we calculate the cross product $\vec{a} \times \vec{c} = (\hat{i} + \hat{j} + \hat{k}) \times (\hat{j} - \hat{k}) = -2\hat{i} + \hat{j} + \hat{k}$. The magnitude of this resulting vector is $\sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{6}$. Substituting these values into our expression gives $l = \frac{2}{\sqrt{6}}$. Squaring this length results in $l^2 = \frac{4}{6} = \frac{2}{3}$, and calculating $3 l^2$ yields $3 \times \frac{2}{3} = 2$.

Since the vector $\mathbf{v}$ lies in the plane defined by $\mathbf{a}$ and $\mathbf{b}$, it can be represented as the linear combination $\mathbf{v} = \lambda\mathbf{a} + \mu\mathbf{b}$. By substituting $\mathbf{a} = 2\hat{i}-\hat{j}+2\hat{k}$ and $\mathbf{b} = \hat{i}+2\hat{j}-\hat{k}$, we express $\mathbf{v}$ as $\mathbf{v} = (2\lambda+\mu)\hat{i} + (2\mu-\lambda)\hat{j} + (2\lambda-\mu)\hat{k}$. Because $\mathbf{v}$ is perpendicular to $3\hat{i}+2\hat{j}-\hat{k}$, the dot product of these two vectors must be zero, which gives $3(2\lambda+\mu) + 2(2\mu-\lambda) - 1(2\lambda-\mu) = 0$. Simplifying this equation results in $2\lambda + 8\mu = 0$, or $\lambda = -4\mu$.

The projection of $\mathbf{v}$ onto $\mathbf{a}$ is given as 19, and given that the magnitude of $\mathbf{a}$ is 3, the dot product $\mathbf{v} \cdot \mathbf{a}$ equals 57. Substituting $\mathbf{v} = \lambda\mathbf{a} + \mu\mathbf{b}$ into this dot product yields $\lambda|\mathbf{a}|^2 + \mu(\mathbf{b} \cdot \mathbf{a}) = 57$. With $|\mathbf{a}|^2 = 9$ and $\mathbf{b} \cdot \mathbf{a} = -2$, we obtain the linear equation $9\lambda - 2\mu = 57$. Substituting $\lambda = -4\mu$ into this equation gives $9(-4\mu) - 2\mu = 57$, which simplifies to $-38\mu = 57$, or $\mu = -3/2$. Consequently, $\lambda = 6$. Using these coefficients, the vector $2\mathbf{v}$ is calculated as $2\mathbf{v} = 12\mathbf{a} - 3\mathbf{b} = 21\hat{i}-18\hat{j}+27\hat{k}$. Finally, the magnitude squared is $|2\mathbf{v}|^2 = 21^2 + (-18)^2 + 27^2 = 441 + 324 + 729 = 1494$.

$|{\vec{P}}|=|{\vec{Q}}|={x}$ .......(i) $|{\vec{P}}+{\vec{Q}}|={n}|{\vec{P}}-{{Q}}^{\to }|$ ${P}^{2}+{Q}^{2}+2 {PQ} \cos \theta$ $={n}^{2}({P}^{2}+{Q}^{2}-2 {P} {Q} \cos \theta )$ Using (i) in above equation $\cos \theta =\frac{{n}^{2}-1}{1+{n}^{2}}$ $\theta =\cos ^{-1}(\frac{{n}^{2}-1}{{n}^{2}+1})$

The core principle to solving this problem lies in leveraging the vector geometric relationship defined by the given magnitude and the property of the cross product. By squaring the given condition $|\vec{c}-\vec{a}|=2\sqrt{2}$, we obtain the equation $|\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{c}\cdot\vec{a}) = 8$. Given that the magnitude $|\vec{a}|$ is 3 and the dot product $\vec{c}\cdot\vec{a}$ is equal to $|\vec{c}|$, substituting these values yields the quadratic expression $|\vec{c}|^2 - 2|\vec{c}| + 1 = 0$, which factors as $(|\vec{c}|-1)^2 = 0$, establishing that $|\vec{c}| = 1$.

We then determine the vector $\vec{a}\times \vec{b}$ by calculating the cross product of $2\hat{i}+\hat{j}-2\hat{k}$ and $\hat{i}+\hat{j}$, which results in $2\hat{i}-2\hat{j}+\hat{k}$. The magnitude of this vector is $\sqrt{2^{2+}(-2)^{2+}1^2} = 3$. Finally, the magnitude of the cross product of this result with $\vec{c}$ is given by the formula $|\vec{u}\times \vec{v}| = |\vec{u}| |\vec{v}| \sin \theta$, where $\theta$ is the angle between the two vectors. Substituting the known magnitudes and the provided angle of $\frac{\pi}{6}$, we calculate the value as $3 \times 1 \times \sin(\frac{\pi}{6})$, which results in $3 \times 1 \times \frac{1}{2} = \frac{3}{2}$.

Because the vectors are mutually perpendicular, their pairwise dot products are zero, which allows the squared magnitude of their sum to be calculated as the sum of their individual squared magnitudes, or $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2$. By assigning each vector a unit magnitude, the resultant vector has a magnitude of $\sqrt{3}$. The angle $\theta$ between any of these vectors and the resultant sum is defined by the dot product $\vec{a} \cdot (\vec{a}+\vec{b}+\vec{c}) = |\vec{a}| |\vec{a}+\vec{b}+\vec{c}| \cos \theta$, which simplifies to $1 = \sqrt{3} \cos \theta$, or $\cos \theta = \frac{1}{\sqrt{3}}$. Utilizing the double angle identity $\cos 2\theta = 2 \cos^2 \theta - 1$, we obtain $\cos 2\theta = 2(\frac{1}{3}) - 1 = -\frac{1}{3}$. Squaring this result and multiplying by 36 yields $36 \left(-\frac{1}{3}\right)^2 = 36 \cdot \frac{1}{9} = 4$.

Since the cross product magnitude $\left| a \times b \right|$ is equal to $\left| a \right|$ and the vectors are perpendicular, the equation $\left| a \right| \left| b \right| \sin 90^{\circ} = \left| a \right|$ simplifies to $\left| b \right| = 1$. Given that $\left| a \right| = \left| b \right|$, it follows that both vectors are unit vectors. To simplify the geometry, one can assign $a = \hat{i}$ and $b = \hat{j}$, which inherently makes their cross product $a \times b = \hat{k}$.

Substituting these assignments, the resultant vector $a + b + (a \times b)$ is represented by the sum $\hat{i} + \hat{j} + \hat{k}$. The angle $\theta$ between this resultant vector and the original vector $a$, which is $\hat{i}$, is determined by the dot product formula $\cos \theta = \frac{ (\hat{i} + \hat{j} + \hat{k}) \cdot \hat{i} }{ \left| \hat{i} + \hat{j} + \hat{k} \right| \left| \hat{i} \right| }$. Computing these values results in $\frac{1}{\sqrt{3} \cdot 1}$, leading to the final angle $\theta = \cos^{-1}\left( \frac{1}{\sqrt{3}} \right)$.

${\vec{A}} \.{{B}}^{\to }=|{{A}}^{\to } \\times {\vec{B}}|$ ${AB} \cos \theta ={AB} \sin \theta \Rightarrow \theta =45°$ $|{{A}}^{\to }-{\vec{B}}|=\\sqrt{{A}^{2}+{B}^{2}-2 {AB} \cos 45°}$ $=\\sqrt{{A}^{2}+{B}^{2}-\\sqrt{2} {AB}}$ Hence option (4).

In the given figure, the magnitude of vectors OA, OB and OC areequal. Let it be m. From figure, we can write {\text{OA}}=m(\cos 30° {{i}}^{\^}+\sin 30° {{j}}^{\^}) =m(\frac{\\sqrt{3}}{2} {{i}}^{\^}+\frac{1}{2} {{j}}^{\^}) {\text{OB}}=m(\cos 60° {{i}}^{\^}-\sin 60°{{j}}^{\^}) =m(\frac{1}{2} {{i}}^{\^}-\frac{\\sqrt{3}}{2} {{j}}^{\^}) {\text{OC}}=m(\cos 45°(-{{i}}^{\^})+\sin 45°{{j}}^{\^}) =m\{\frac{-1}{\\sqrt{2}} {{i}}^{\^}+\frac{1}{\\sqrt{2}} {{j}}^{\^}\} According to the question,{\text{OA}}+{\text{OB}}-{\text{OC}}=m(\frac{\\sqrt{3}}{2} {{i}}^{\^}+\frac{1}{2} {{j}}^{\^})+m(\frac{1}{2} {{i}}^{\^}-\frac{\\sqrt{3}}{2} {{j}}^{\^}) -m(-\frac{1}{\\sqrt{2}} {{i}}^{\^}+\frac{1}{\\sqrt{2}} {{j}}^{\^}) =m {{i}}^{\^}(\frac{\\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{\\sqrt{2}})+m {{j}}^{\^}(\frac{1}{2}-\frac{\\sqrt{3}}{2}-\frac{1}{\\sqrt{2}}) ∴ Angle with X - axis is given by $\tan \theta =\frac{y}{x}={m(\frac{1}{2}-\frac{\\sqrt{3}}{2}-\frac{1}{\\sqrt{2}})}/{m(\frac{\\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{\\sqrt{2}})}$ $\Rightarrow \tan \theta ={\frac{1}{2}-\frac{\\sqrt{3}}{2}-\frac{\\sqrt{2}}{2}}/{\frac{\\sqrt{3}}{2}+\frac{1}{2}+\frac{\\sqrt{2}}{2}}$ $\Rightarrow \tan \theta =\frac{1-\\sqrt{3}-\sqrt{2}}{\sqrt{3}+1+\\sqrt{2}}$ $\Rightarrow \theta =\tan ^{-1}(\frac{1-\\sqrt{3}-\\sqrt{2}}{1+\\sqrt{3}+\\sqrt{2}})$

A vector that bisects the angle between two other vectors must lie along the direction of the sum or the difference of their respective unit vectors. By normalizing $\vec{b}=\vec{i}+\vec{j}$ and $\vec{c}=\vec{i}-\vec{j}+4\vec{k}$, we identify the unit vectors as $\hat{b} = \frac{\vec{i}+\vec{j}}{\sqrt{2}}$ and $\hat{c} = \frac{\vec{i}-\vec{j}+4\vec{k}}{3\sqrt{2}}$. Any vector $\vec{a}$ bisecting the angle between them follows the form $\vec{a} = \lambda(\hat{b} \pm \hat{c})$.

Evaluating the sum of the unit vectors leads to $\vec{a} = \lambda \left( \frac{3\vec{i}+3\vec{j} + \vec{i}-\vec{j}+4\vec{k}}{3\sqrt{2}} \right)$, which simplifies to $\frac{\lambda}{3\sqrt{2}}(4\vec{i}+2\vec{j}+4\vec{k})$. To align the $y$-component of this expression with the $2\vec{j}$ term in the definition of $\vec{a}$, we set the coefficient of $\vec{j}$ equal to $2$, resulting in $\frac{2\lambda}{3\sqrt{2}} = 2$ and yielding $\vec{a} = 4\vec{i}+2\vec{j}+4\vec{k}$. As this vector does not align with the given options, the alternative angle bisector must be considered.

Examining the difference of the unit vectors, we obtain $\vec{a} = \mu \left( \frac{3\vec{i}+3\vec{j} - (\vec{i}-\vec{j}+4\vec{k})}{3\sqrt{2}} \right)$, which simplifies to $\frac{\mu}{3\sqrt{2}}(2\vec{i}+4\vec{j}-4\vec{k})$. Matching the $y$-component to $2\vec{j}$ requires $\frac{4\mu}{3\sqrt{2}} = 2$, which leads to $\frac{\mu}{3\sqrt{2}} = \frac{1}{2}$. Substituting this constant back into the vector expression gives $\vec{a} = \vec{i}+2\vec{j}-2\vec{k}$. Calculating the dot product of this vector with $\vec{k}$ provides $-2$, demonstrating that $\vec{a}\cdot\vec{k}+2=0$.

The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ represents the volume-related function $f(x) = x^3 - 27x + 26$, which is derived from the determinant of the matrix formed by the given vectors. To find the local maxima, we identify the critical points by setting the derivative $f'(x) = 3x^2 - 27$ to zero, yielding $x = \pm 3$. Applying the second derivative test, where $f''(x) = 6x$, reveals that the function attains a local maximum at $x_0 = -3$ because the second derivative is negative at that point.

Substituting $x_0 = -3$ into the original vector definitions gives $\vec{a} = -3 \hat{i} - 2 \hat{j} + 3 \hat{k}$, $\vec{b} = -2 \hat{i} - 3 \hat{j} - \hat{k}$, and $\vec{c} = 7 \hat{i} - 2 \hat{j} - 3 \hat{k}$. Calculating the individual dot products results in $\vec{a} \cdot \vec{b} = 9$, $\vec{b} \cdot \vec{c} = -5$, and $\vec{c} \cdot \vec{a} = -26$. Summing these three values provides the final result of $9 - 5 - 26$, which equals $-22$.

The squared magnitude of the vector expression $\left|\overset{\rightarrow}{a}+\overset{\rightarrow}{b}-\overset{\rightarrow}{c}\right|$ provides a direct path to the solution by expanding the dot product. Given $\left|\overset{\rightarrow}{a}\right|=2$, $\left|\overset{\rightarrow}{b}\right|=4$, and $\left|\overset{\rightarrow}{c}\right|=4$, expanding the squared expression yields $\left|\overset{\rightarrow}{a}\right|^2 + \left|\overset{\rightarrow}{b}\right|^2 + \left|\overset{\rightarrow}{c}\right|^2 + 2\overset{\rightarrow}{a}\cdot\overset{\rightarrow}{b} - 2\overset{\rightarrow}{a}\cdot\overset{\rightarrow}{c} - 2\overset{\rightarrow}{b}\cdot\overset{\rightarrow}{c}$. Since the projections of $\overset{\rightarrow}{b}$ and $\overset{\rightarrow}{c}$ onto $\overset{\rightarrow}{a}$ are equal, $\overset{\rightarrow}{a}\cdot\overset{\rightarrow}{b} = \overset{\rightarrow}{a}\cdot\overset{\rightarrow}{c}$, which causes the terms $2\overset{\rightarrow}{a}\cdot\overset{\rightarrow}{b} - 2\overset{\rightarrow}{a}\cdot\overset{\rightarrow}{c}$ to cancel out to zero. Given that $\overset{\rightarrow}{b}$ and $\overset{\rightarrow}{c}$ are perpendicular, the term $\overset{\rightarrow}{b}\cdot\overset{\rightarrow}{c}$ is also zero, simplifying the entire expression to the sum of the squared magnitudes, resulting in $4 + 16 + 16 = 36$. Taking the square root of this result yields 6.

Squaring the sum $\overset{\rightarrow}{a} + \overset{\rightarrow}{b} + \overset{\rightarrow}{c} = 0$ provides an immediate path to finding $\lambda$. With each vector having a magnitude of $1$, the expansion $|\overset{\rightarrow}{a} + \overset{\rightarrow}{b} + \overset{\rightarrow}{c}|^2 = |\overset{\rightarrow}{a}|^2 + |\overset{\rightarrow}{b}|^2 + |\overset{\rightarrow}{c}|^2 + 2(\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{b} + \overset{\rightarrow}{b} \cdot \overset{\rightarrow}{c} + \overset{\rightarrow}{c} \cdot \overset{\rightarrow}{a}) = 0$ leads directly to $1 + 1 + 1 + 2\lambda = 0$, confirming $\lambda = -\frac{3}{2}$.

Cross products behave under the same constraint $\overset{\rightarrow}{a} + \overset{\rightarrow}{b} + \overset{\rightarrow}{c} = 0$. By crossing this equation with $\overset{\rightarrow}{a}$, one finds $\overset{\rightarrow}{a} \times \overset{\rightarrow}{b} = \overset{\rightarrow}{c} \times \overset{\rightarrow}{a}$. Cyclic symmetry further dictates that $\overset{\rightarrow}{a} \times \overset{\rightarrow}{b} = \overset{\rightarrow}{b} \times \overset{\rightarrow}{c} = \overset{\rightarrow}{c} \times \overset{\rightarrow}{a}$. Substituting these equal cross products into the expression for $\overset{\rightarrow}{d}$ results in $\overset{\rightarrow}{d} = 3(\overset{\rightarrow}{a} \times \overset{\rightarrow}{b})$, completing the ordered pair.

We can determine the behavior of this expression by relating the magnitudes of the sum and difference of two unit vectors, $\vec{a}$ and $\vec{b}$, to the angle $\theta$ between them. Using the properties of vector addition and subtraction, $|\vec{a}+\vec{b}|$ is equal to $2|\cos(\frac{\theta}{2})|$ and $|\vec{a}-\vec{b}|$ is equal to $2|\sin(\frac{\theta}{2})|$. Substituting these geometric identities into the given expression yields $2\sqrt{3}|\cos(\frac{\theta}{2})| + 2|\sin(\frac{\theta}{2})|$. This expression follows the form $A\cos x + B\sin x$, which is bounded by a maximum value of $\sqrt{A^{2+}B^2}$. Identifying $A$ as $2\sqrt{3}$ and $B$ as $2$, we compute the maximum as $\sqrt{(2\sqrt{3})^2 + 2^2}$, which simplifies to $\sqrt{12+4}$ and results in 4.

We determine the magnitude of the vector $\overset{\rightarrow}{c}$ by utilizing the dot product formula $\overset{\rightarrow}{b} \cdot \overset{\rightarrow}{c} = \left|\overset{\rightarrow}{b}\right| \left|\overset{\rightarrow}{c}\right| \cos\left(\frac{\pi}{3}\right)$. Given $\overset{\rightarrow}{b} \cdot \overset{\rightarrow}{c} = 10$ and $\left|\overset{\rightarrow}{b}\right| = 5$, the equation $5 \left|\overset{\rightarrow}{c}\right| \left(\frac{1}{2}\right) = 10$ confirms that $\left|\overset{\rightarrow}{c}\right| = 4$.

The magnitude of the cross product $\overset{\rightarrow}{b} \times \overset{\rightarrow}{c}$ is given by $\left|\overset{\rightarrow}{b}\right| \left|\overset{\rightarrow}{c}\right| \sin\left(\frac{\pi}{3}\right)$, which simplifies to $5 \cdot 4 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3}$. Because the vector $\overset{\rightarrow}{a}$ is perpendicular to $\overset{\rightarrow}{b} \times \overset{\rightarrow}{c}$, the angle between them is $\frac{\pi}{2}$, which makes the sine component of the resulting cross product magnitude equal to 1. Thus, the final magnitude $\left|\overset{\rightarrow}{a} \times \left(\overset{\rightarrow}{b} \times \overset{\rightarrow}{c}\right)\right|$ is calculated as the product $\left|\overset{\rightarrow}{a}\right| \cdot \left|\overset{\rightarrow}{b} \times \overset{\rightarrow}{c}\right|$, which results in $\sqrt{3} \cdot 10\sqrt{3} = 30$.

The volume of a parallelopiped defined by its coterminous edges is equal to the absolute value of the scalar triple product of the vectors forming those edges, which is computed as the determinant of the matrix containing their components. Setting the determinant of the vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$ equal to $\pm 1$ yields the equation $3 - \lambda = \pm 1$, which confirms that the possible values for $\lambda$ are 2 and 4.

Once these values are established, the angle $\theta$ between the edges $\vec{u}$ and $\vec{w}$ is determined using the geometric definition $\cos\theta = \frac{\vec{u} \cdot \vec{w}}{|\vec{u}| |\vec{w}|}$. Substituting $\lambda = 2$ into the expression leads to $\cos\theta = \frac{5}{6}$, while substituting $\lambda = 4$ yields $\cos\theta = \frac{7}{\sqrt{18}\sqrt{6}}$, which simplifies to $\frac{7}{6\sqrt{3}}$.

To determine if two lines intersect, we equate their respective $\hat{i}$, $\hat{j}$, and $\hat{k}$ components to check if a common pair of parameters $l$ and $m$ exists. Equating these components yields the system $1+2l = 2+m$, $-1 = m-1$, and $l = -m$. Solving the second equation gives $m = 0$, and substituting this result into the third equation determines that $l = 0$. Testing these values in the first equation shows that the left side equals 1 while the right side equals 2, confirming that the lines do not intersect for any values of $l$ and $m$.

The vector triple product identity $\vec{p} \times (\vec{q} \times \vec{r}) = \vec{q}(\vec{p} \cdot \vec{r}) - \vec{r}(\vec{p} \cdot \vec{q})$ reveals that any expression of the form $\hat{u} \times (\vec{a} \times \hat{u})$, where $\hat{u}$ is a unit vector, simplifies to $\vec{a} - (\vec{a} \cdot \hat{u})\hat{u}$. Squaring the magnitude of this result yields $|\vec{a}|^2 - (\vec{a} \cdot \hat{u})^2$, which represents the square of the component of $\vec{a}$ perpendicular to the chosen axis.