Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

The cross product equality $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ can be rearranged into $(\vec{r} - \vec{c}) \times \vec{b} = \vec{0}$. This form signifies that the vector $\vec{r} - \vec{c}$ must be parallel to $\vec{b}$, which allows us to express the unknown vector as $\vec{r} = \vec{c} + \lambda \vec{b}$ for some scalar constant $\lambda$. Substituting this expression into the dot product condition $\vec{r} \cdot \vec{a} = 0$ gives $(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 0$, which expands to $\vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$.

Using the provided vectors $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$, and $\vec{c} = 5\hat{i} - 3\hat{j} + 3\hat{k}$, we calculate the dot products as $\vec{c} \cdot \vec{a} = (5)(1) + (-3)(2) + (3)(3) = 5 - 6 + 9 = 8$ and $\vec{b} \cdot \vec{a} = (1)(1) + (-1)(2) + (2)(3) = 1 - 2 + 6 = 5$. Substituting these values into our linear equation yields $8 + 5\lambda = 0$, which means $\lambda = -8/5$.

We then determine the vector $\vec{r}$ by substituting $\lambda$ back into $\vec{r} = \vec{c} + \lambda \vec{b}$, resulting in $\vec{r} = (5\hat{i} - 3\hat{j} + 3\hat{k}) - \frac{8}{5}(\hat{i} - \hat{j} + 2\hat{k})$. Simplifying the components gives $\vec{r} = \frac{1}{5}(25\hat{i} - 15\hat{j} + 15\hat{k} - 8\hat{i} + 8\hat{j} - 16\hat{k}) = \frac{1}{5}(17\hat{i} - 7\hat{j} - \hat{k})$. The square of the magnitude is then $|\vec{r}|^2 = \frac{1}{25}(17^2 + (-7)^2 + (-1)^2) = \frac{289 + 49 + 1}{25} = \frac{339}{25}$. Multiplying this result by 25 gives $25|\vec{r}|^2 = 339$.

When three coplanar, concurrent vectors share the same angle between any two of them, they are oriented at $120^\circ$ relative to one another. Using the vector identity $(\vec{u} \times \vec{v}) \cdot (\vec{w} \times \vec{z}) = (\vec{u} \cdot \vec{w})(\vec{v} \cdot \vec{z}) - (\vec{u} \cdot \vec{z})(\vec{v} \cdot \vec{w})$, we can simplify each term in the given expression. The dot product of any two vectors $\vec{u}$ and $\vec{v}$ at $120^\circ$ is given by
$\vec{u} \cdot \vec{v} = -0.5|\vec{u}||\vec{v}|$
where $|\vec{u}|$ and $|\vec{v}|$ denote their respective magnitudes.

Applying this identity to the first term $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c})$ yields $(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{b})$. Substituting the dot products results in $(-0.5|\vec{a}||\vec{b}|)(-0.5|\vec{b}||\vec{c}|) - (-0.5|\vec{a}||\vec{c}|)|\vec{b}|^2$, which simplifies to $0.75|\vec{a}||\vec{b}|^2|\vec{c}|$. Summing this with the similarly derived values for the other two terms, $0.75|\vec{a}||\vec{b}||\vec{c}|^2$ and $0.75|\vec{a}|^2|\vec{b}||\vec{c}|$, results in the overall expression
$0.75|\vec{a}||\vec{b}||\vec{c}|(|\vec{a}| + |\vec{b}| + |\vec{c}|) = 168$
Given that the product $|\vec{a}||\vec{b}||\vec{c}| = 14$, we set up $0.75(14)(|\vec{a}| + |\vec{b}| + |\vec{c}|) = 168$, which simplifies to $10.5(|\vec{a}| + |\vec{b}| + |\vec{c}|) = 168$, identifying the sum as 16.

The area of a triangle defined by two vectors is half the magnitude of their cross product, while the area of a parallelogram $S$ spanned by $\vec{a}$ and $\vec{b}$ is equal to $|\vec{a} \times \vec{b}|$. Splitting the quadrilateral $OABC$ into two triangles, $\triangle OAB$ and $\triangle OBC$, allows us to calculate their individual areas using this principle. The area of $\triangle OAB$ is defined as $\frac{1}{2} |\vec{OA} \times \vec{OB}|$, which becomes $\frac{1}{2} |\vec{a} \times (12\vec{a} + 4\vec{b})|$. Since the cross product of any vector with itself is zero, this simplifies to $\frac{1}{2} |4(\vec{a} \times \vec{b})|$, yielding $2 |\vec{a} \times \vec{b}|$.

Similarly, the area of $\triangle OBC$ is determined by $\frac{1}{2} |\vec{OB} \times \vec{OC}|$, or $\frac{1}{2} |(12\vec{a} + 4\vec{b}) \times \vec{b}|$. Distributing the cross product results in $\frac{1}{2} |12(\vec{a} \times \vec{b}) + 4(\vec{b} \times \vec{b})|$, which reduces to $6 |\vec{a} \times \vec{b}|$ because $\vec{b} \times \vec{b} = 0$. Adding these two triangle areas, $2 |\vec{a} \times \vec{b}| + 6 |\vec{a} \times \vec{b}|$, gives a total area of $8 |\vec{a} \times \vec{b}|$. Dividing this total area by the area of the parallelogram $S$, which is $|\vec{a} \times \vec{b}|$, produces a ratio of 8.

In a triangle $ABC$, the sides follow the vector addition rule $\overrightarrow{CB} = \overrightarrow{CA} + \overrightarrow{AB}$. Given $\overrightarrow{CA} = 4\hat{i} + 3\hat{j} + \delta\hat{k}$ and $\overrightarrow{AB} = -2\hat{i} + \hat{j} + 3\hat{k}$, summing these vectors yields $\overrightarrow{CB} = 2\hat{i} + 4\hat{j} + (\delta+3)\hat{k}$. The area of this triangle is given by the formula $\frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|$. Recalling that $\overrightarrow{AC} = -4\hat{i} - 3\hat{j} - \delta\hat{k}$, we determine the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$ by calculating the determinant formed by the basis vectors and the components of $\overrightarrow{AB}$ and $\overrightarrow{AC}$, which results in $(9-\delta)\hat{i} - (2\delta+12)\hat{j} + 10\hat{k}$.

Setting half the magnitude of this cross product equal to $5\sqrt{6}$ implies that the magnitude itself is $10\sqrt{6}$, so the square of the magnitude must equal $600$. This leads to the equation $(9-\delta)^2 + (2\delta+12)^2 + 10^2 = 600$, which expands and simplifies to $5\delta^2 + 30\delta - 275 = 0$, or $\delta^2 + 6\delta - 55 = 0$. Factoring the quadratic gives $(\delta+11)(\delta-5) = 0$, and since the problem specifies $\delta > 0$, we find that $\delta = 5$. Substituting this value back into the vectors, we have $\overrightarrow{CB} = 2\hat{i} + 4\hat{j} + 8\hat{k}$ and $\overrightarrow{CA} = 4\hat{i} + 3\hat{j} + 5\hat{k}$. The final dot product $\overrightarrow{CB} \cdot \overrightarrow{CA}$ is calculated as $(2)(4) + (4)(3) + (8)(5)$, which equals $60$.

Collinearity implies that vectors can be expressed as scalar multiples of one another, and a linear combination of linearly independent vectors equaling the zero vector requires each individual coefficient to be zero. The conditions specify $\vec{a} + 5\vec{b} = k_1\vec{c}$ and $\vec{b} + 6\vec{c} = k_2\vec{a}$ for some scalars $k_1$ and $k_2$. Substituting the expression for $\vec{a} = k_1\vec{c} - 5\vec{b}$ into the second equation yields $\vec{b} + 6\vec{c} = k_2(k_1\vec{c} - 5\vec{b})$, which simplifies to $(1 + 5k_2)\vec{b} + (6 - k_1 k_2)\vec{c} = \vec{0}$.

Because $\vec{b}$ and $\vec{c}$ are non-collinear, their coefficients must vanish, leading to the system $1 + 5k_2 = 0$ and $6 - k_1 k_2 = 0$. Solving these equations gives $k_2 = -1/5$ and $k_1 = -30$. Substituting $k_1 = -30$ back into the first expression results in $\vec{a} + 5\vec{b} = -30\vec{c}$, which rearranges to $\vec{a} + 5\vec{b} + 30\vec{c} = \vec{0}$. Matching this with the form $\vec{a} + \alpha\vec{b} + \beta\vec{c} = \vec{0}$ identifies $\alpha = 5$ and $\beta = 30$, resulting in a sum $\alpha + \beta$ of $35$.

The condition $(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = \vec{0}$ implies that the sum $\vec{a} + \vec{b}$ must be parallel to $\vec{c}$, because the cross product of any vector with itself is the zero vector. Given $\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$, their sum is $(\lambda + 3)\hat{i} + \hat{k}$. Consequently, we can express $\vec{c}$ as a scalar multiple $k$ of this sum, represented as $\vec{c} = k((\lambda + 3)\hat{i} + \hat{k})$.

Using the dot product relations, $\vec{a} \cdot \vec{c} = -17$ leads to $k(\lambda(\lambda + 3) - 1) = -17$, while $\vec{b} \cdot \vec{c} = -20$ yields $k(3(\lambda + 3) + 2) = -20$. Simplifying the second equation gives $k(3\lambda + 11) = -20$, from which we isolate $k$ as $\frac{-20}{3\lambda + 11}$. Substituting this into the first equation produces the quadratic relationship $20\lambda^2 + 9\lambda - 207 = 0$. Given that $\lambda$ must be an integer, solving this quadratic reveals $\lambda = 3$, which results in a scalar $k = -1$.

With $\lambda = 3$ and $k = -1$, the vector $\vec{c}$ becomes $-6\hat{i} - \hat{k}$. We compute the requested cross product with the vector $3\hat{i} + \hat{j} + \hat{k}$ by evaluating the determinant of the matrix formed by their components, resulting in $\hat{i} + 3\hat{j} - 6\hat{k}$. Squaring the magnitude of this resultant vector involves summing the squares of its components, $1^2 + 3^2 + (-6)^2$, yielding 46.

Because each vector is equal to the cross product of the other two, the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are mutually orthogonal, meaning their mutual dot products are zero. Applying the magnitude property of the cross product $|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin 90^\circ$, the given equations convert into the scalar system $|\vec{a}| |\vec{b}| = 4 |\vec{c}|$, $|\vec{b}| |\vec{c}| = 9 |\vec{a}|$, and $|\vec{c}| |\vec{a}| = \alpha |\vec{b}|$. Multiplying these three relations together yields the product $(|\vec{a}| |\vec{b}| |\vec{c}|)^2 = 36 \alpha (|\vec{a}| |\vec{b}| |\vec{c}|)$, which simplifies to $|\vec{a}| |\vec{b}| |\vec{c}| = 36 \alpha$.

By dividing this product by the original magnitude relations, we isolate the individual magnitudes: $|\vec{c}|^2 = 9 \alpha$ so $|\vec{c}| = 3 \sqrt{\alpha}$, $|\vec{a}|^2 = 4 \alpha$ so $|\vec{a}| = 2 \sqrt{\alpha}$, and $|\vec{b}|^2 = 36$ so $|\vec{b}| = 6$. Substituting these values into the sum condition yields the relation $2 \sqrt{\alpha} + 6 + 3 \sqrt{\alpha} = 36$, which leads directly to $\sqrt{\alpha} = 6$ and subsequently $\alpha = 36$.

The magnitude squared of the difference between two vectors $\vec{u}$ and $\vec{v}$ is determined by the algebraic identity $|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2(\vec{u} \cdot \vec{v})$. A fundamental geometric property of the cross product is that the vector resulting from $\vec{b} \times \vec{a}$ is always perpendicular to the vector $\vec{b}$. Because the dot product of any two orthogonal vectors is zero, the term involving the dot product in our expansion vanishes, which simplifies the expression significantly.

Expanding the original problem using this identity results in $|\vec{b} \times \vec{a}|^2 + |\vec{b}|^2 - 2(\vec{b} \times \vec{a} \cdot \vec{b})$. Since the dot product term is 0, the equation reduces to the sum of the squares of the magnitudes, $|\vec{b} \times \vec{a}|^2 + |\vec{b}|^2$. Substituting the given values where the magnitude of the cross product is 2 and the magnitude of $\vec{b}$ is 1 leads to $2^2 + 1^2$. Calculating these values yields $4 + 1$, which equals 5.

The vector triple product identity dictates that $\vec{a} \times (\vec{b} \times \vec{c})$ can be expanded as $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$. By equating this expansion to the given expression $\frac{\sqrt{3}}{2}\vec{b} + \frac{\sqrt{3}}{2}\vec{c}$ and noting that $\vec{b}$ and $\vec{c}$ are non-parallel, we can directly equate their respective scalar coefficients to obtain $\vec{a} \cdot \vec{c} = \frac{\sqrt{3}}{2}$ and $\vec{a} \cdot \vec{b} = -\frac{\sqrt{3}}{2}$. Because both $\vec{a}$ and $\vec{b}$ are unit vectors, the dot product simplifies to $\cos\theta = -\frac{\sqrt{3}}{2}$. Solving this equation for $\theta$ within the standard range yields an angle of $\frac{5\pi}{6}$.

The vector equation $\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$ implies that the vector $(\vec{r} - \vec{c})$ must be collinear with $\vec{a}$, allowing us to express the unknown vector as $\vec{r} = \vec{c} + \lambda\vec{a}$ for some scalar constant $\lambda$. Imposing the orthogonal constraint $\vec{r} \cdot \vec{b} = 0$ requires that $(\vec{c} + \lambda\vec{a}) \cdot \vec{b} = 0$, which simplifies to the linear relation $\vec{c} \cdot \vec{b} + \lambda(\vec{a} \cdot \vec{b}) = 0$.

Calculating the dot products using the given components $\vec{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$, $\vec{b} = \hat{i} + \hat{k}$, and $\vec{c} = \hat{i} + 2\hat{j} - 3\hat{k}$, we find $\vec{c} \cdot \vec{b} = -2$ and $\vec{a} \cdot \vec{b} = 7$. Substituting these into our equation $-2 + 7\lambda = 0$ identifies the scalar as $\lambda = \frac{2}{7}$. Substituting this value back into the expression for $\vec{r}$ yields $\vec{r} = (\hat{i} + 2\hat{j} - 3\hat{k}) + \frac{2}{7}(2\hat{i} - 7\hat{j} + 5\hat{k})$, which simplifies to $\frac{1}{7}(11\hat{i} - 11\hat{k})$. The magnitude $|\vec{r}|$ is then found by evaluating $\frac{1}{7}\sqrt{11^2 + 0^2 + (-11)^2}$, resulting in $\frac{11}{7} \sqrt{2}$.

The scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is defined as $\vec{u} \cdot (\vec{v} \times \vec{w})$, which equates to the product of the magnitudes $|\vec{u}|$ and $|\vec{v} \times \vec{w}|$ multiplied by the cosine of the angle between them. To achieve a negative minimum value, the vectors $\vec{u}$ and $\vec{v} \times \vec{w}$ must point in opposite directions, which sets the cosine of the angle to $-1$.

To evaluate the cross product, we calculate the determinant of the matrix formed by the unit vectors and the given components:
$\vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -3 \\ 2\alpha & 1 & -1 \end{vmatrix} = \hat{i}(-2 + 3) - \hat{j}(-\alpha + 6\alpha) + \hat{k}(\alpha - 4\alpha) = \hat{i} - 5\alpha \hat{j} - 3\alpha \hat{k}$
The magnitude of this cross product is $\sqrt{1 + (-5\alpha)^2 + (-3\alpha)^2} = \sqrt{1 + 34\alpha^2}$. Equating the minimum value $-|\vec{u}| |\vec{v} \times \vec{w}|$ to the provided value $-\alpha \sqrt{3401}$ leads to $\alpha \sqrt{1 + 34\alpha^2} = \alpha \sqrt{3401}$, which implies $1 + 34\alpha^2 = 3401$. Solving this equation yields $34\alpha^2 = 3400$, so $\alpha^2 = 100$ and $\alpha = 10$.

Because $\vec{u}$ is antiparallel to $\vec{v} \times \vec{w}$, we can represent it as $\vec{u} = \lambda(\hat{i} - 5\alpha \hat{j} - 3\alpha \hat{k})$ for some scalar $\lambda$. Using $|\vec{u}| = \alpha$, we square the magnitude to get $|\vec{u}|^2 = \lambda^2(1 + 25\alpha^2 + 9\alpha^2) = \alpha^2$. Substituting $\alpha^2 = 100$ results in $\lambda^2(1 + 34(100)) = 100$, which simplifies to $\lambda^2(3401) = 100$, or $\lambda^2 = \frac{100}{3401}$. Since $\vec{u} \cdot \hat{i}$ is simply $\lambda$, the value of $(\vec{u} \cdot \hat{i})^2$ is $\frac{100}{3401}$. Identifying $m = 100$ and $n = 3401$ as coprime natural numbers, the sum $m + n$ is $3501$.

Expanding the vector triple product using the identity $(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$ allows us to equate this to the given expression, leading to $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} = \frac{1}{3}|\vec{b}||\vec{c}|\vec{a}$. Rearranging this equality reveals $(\vec{a} \cdot \vec{c})\vec{b} = (\vec{b} \cdot \vec{c} + \frac{1}{3}|\vec{b}||\vec{c}|)\vec{a}$. Because $\vec{a}$ and $\vec{b}$ are non-collinear, this relationship can only be true if the scalar coefficients on both sides are zero; specifically, $\vec{b} \cdot \vec{c} + \frac{1}{3}|\vec{b}||\vec{c}| = 0$. Using the definition of the dot product $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}| \cos \theta$, the equation simplifies to $|\vec{b}||\vec{c}| \cos \theta = -\frac{1}{3}|\vec{b}||\vec{c}|$, which gives $\cos \theta = -\frac{1}{3}$. Applying the trigonometric identity $\sin \theta = \sqrt{1 - \cos^2 \theta}$ yields $\sin \theta = \sqrt{1 - (-\frac{1}{3})^2}$, resulting in a value of $\frac{2\sqrt{2}}{3}$.

The decomposition of vector $\vec{b}$ into components $\vec{b}_1$ parallel to $\vec{a}$ and $\vec{b}_2$ perpendicular to $\vec{a}$ is given by $\vec{b}_1 = \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \vec{a}$ and $\vec{b}_2 = \vec{b} - \vec{b}_1$. Given $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = 3\hat{j} + 4\hat{k}$, the projection scalar is calculated as $\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} = \frac{(3\hat{j} + 4\hat{k}) \cdot (\hat{i} + \hat{j})}{1^2 + 1^2} = \frac{3}{2}$. Consequently, the parallel component is $\vec{b}_1 = \frac{3}{2}(\hat{i} + \hat{j}) = \frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}$, and the perpendicular component is $\vec{b}_2 = (3\hat{j} + 4\hat{k}) - (\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}) = -\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j} + 4\hat{k}$.

The cross product is computed as $\vec{b}_1 \times \vec{b}_2 = (\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}) \times (-\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j} + 4\hat{k})$. Expanding this expression yields $\frac{3}{2} [ \hat{i} \times (-\frac{3}{2}\hat{i}) + \hat{i} \times (\frac{3}{2}\hat{j}) + \hat{i} \times 4\hat{k} + \hat{j} \times (-\frac{3}{2}\hat{i}) + \hat{j} \times (\frac{3}{2}\hat{j}) + \hat{j} \times 4\hat{k} ]$. Using the fundamental cross product relations $\hat{i} \times \hat{j} = \hat{k}$, $\hat{i} \times \hat{k} = -\hat{j}$, $\hat{j} \times \hat{i} = -\hat{k}$, and $\hat{j} \times \hat{k} = \hat{i}$, the terms simplify to $\frac{3}{2} [ 0 + \frac{3}{2}\hat{k} - 4\hat{j} + \frac{3}{2}\hat{k} + 0 + 4\hat{i} ]$, which results in $\frac{3}{2} (4\hat{i} - 4\hat{j} + 3\hat{k}) = 6\hat{i} - 6\hat{j} + \frac{9}{2}\hat{k}$.

The area of a parallelogram defined by adjacent vectors $\vec{a}$ and $\vec{b}$ is equal to the magnitude of their cross product. Calculating the cross product leads to the following expression:
$\vec{a} \times \vec{b} = (2+\alpha)\hat{i} - (\alpha-2)\hat{j} + (\alpha^{2+}4)\hat{k}$
Squaring the magnitude of this result and equating it to the square of the given area $15(\alpha^{2+}4)$ produces the algebraic equation $(2+\alpha)^{2+}(\alpha-2)^{2+}(\alpha^{2+}4)^2 = 15(\alpha^{2+}4)$. Simplifying this expansion results in $\alpha^4 + 10\alpha^2 + 24 = 15\alpha^2 + 60$, which simplifies further to $\alpha^4 - 5\alpha^2 - 36 = 0$. Factoring this expression as $(\alpha^{2-}9)(\alpha^{2+}4)=0$ confirms that $\alpha^2 = 9$, as $\alpha^2$ must be a positive value.

With $\alpha^2 = 9$ established, the squared magnitudes of the vectors are determined as $|\vec{a}|^2 = \alpha^2 + 2^2 + (-1)^2 = 9+4+1 = 14$ and $|\vec{b}|^2 = (-2)^2 + \alpha^2 + 1^2 = 4+9+1 = 14$. The dot product of the vectors is $\vec{a} \cdot \vec{b} = (\alpha)(-2) + (2)(\alpha) + (-1)(1) = -1$. Substituting these values into the required expression $2|\vec{a}|^2 + (\vec{a}\cdot\vec{b})|\vec{b}|^2$ yields $2(14) + (-1)(14)$, which simplifies to 14.

The vector triple product identity states that for any three vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$, the expression $(\vec{u} \times \vec{v}) \times \vec{w}$ can be simplified as $(\vec{u} \cdot \vec{w}) \vec{v} - (\vec{v} \cdot \vec{w}) \vec{u}$. Applying this to $((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})) \times (\vec{a} - \vec{b})$, we let $\vec{u} = (\vec{a} + \vec{b})$, $\vec{v} = (\vec{a} \times \vec{b})$, and $\vec{w} = (\vec{a} - \vec{b})$. Because the vector $\vec{a} \times \vec{b}$ is orthogonal to both $\vec{a}$ and $\vec{b}$, the dot product $(\vec{a} \times \vec{b}) \cdot (\vec{a} - \vec{b})$ is zero. This simplifies the expression to $((\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})) (\vec{a} \times \vec{b})$, which is $(|\vec{a}|^2 - |\vec{b}|^2) (\vec{a} \times \vec{b})$.