Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

In any triangle, the centroid, orthocentre, and circumcentre are collinear, forming the Euler line, where the centroid divides the line segment connecting the circumcentre and the orthocentre in a $2:1$ ratio. By defining the position vector of the centroid $\vec{G}$ as the average of the three vertex vectors, we find $\vec{G} = \frac{(4\vec{p}+\vec{q}-3\vec{r}) + (-5\vec{p}+\vec{q}+2\vec{r}) + (2\vec{p}-\vec{q}+2\vec{r})}{3}$, which simplifies to $\frac{\vec{p}+\vec{q}+\vec{r}}{3}$.

Given that the orthocentre $\vec{H}$ is $\frac{\vec{p}+\vec{q}+\vec{r}}{4}$, we apply the property $\vec{G} = \frac{2\vec{O} + \vec{H}}{3}$ where $\vec{O} = \alpha\vec{p}+\beta\vec{q}+\gamma\vec{r}$ represents the circumcentre. Substituting these expressions into the equation yields $\frac{\vec{p}+\vec{q}+\vec{r}}{3} = \frac{2(\alpha\vec{p}+\beta\vec{q}+\gamma\vec{r}) + \frac{\vec{p}+\vec{q}+\vec{r}}{4}}{3}$. Multiplying the entire equation by 3 gives $\vec{p}+\vec{q}+\vec{r} = 2(\alpha\vec{p}+\beta\vec{q}+\gamma\vec{r}) + \frac{1}{4}(\vec{p}+\vec{q}+\vec{r})$, and subtracting $\frac{1}{4}(\vec{p}+\vec{q}+\vec{r})$ from both sides leads to $\frac{3}{4}(\vec{p}+\vec{q}+\vec{r}) = 2(\alpha\vec{p}+\beta\vec{q}+\gamma\vec{r})$. Dividing by 2 establishes that $\alpha = \frac{3}{8}$, $\beta = \frac{3}{8}$, and $\gamma = \frac{3}{8}$. Consequently, calculating $\alpha+2\beta+5\gamma$ results in $\frac{3}{8} + 2\left(\frac{3}{8}\right) + 5\left(\frac{3}{8}\right)$, which equals $\frac{3+6+15}{8} = 3$.

Two vectors are perpendicular if and only if their dot product equals zero. Expanding the dot product of the expressions $\lambda\overset{\rightarrow}{a}+2\overset{\rightarrow}{b}$ and $3\overset{\rightarrow}{a}-\lambda\overset{\rightarrow}{b}$ gives $3\lambda(\overset{\rightarrow}{a}\cdot\overset{\rightarrow}{a}) - \lambda^2(\overset{\rightarrow}{a}\cdot\overset{\rightarrow}{b}) + 6(\overset{\rightarrow}{b}\cdot\overset{\rightarrow}{a}) - 2\lambda(\overset{\rightarrow}{b}\cdot\overset{\rightarrow}{b}) = 0$. Given that $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$ are unit vectors with an angle of $\frac{\pi}{3}$ between them, their dot product is $\overset{\rightarrow}{a}\cdot\overset{\rightarrow}{b} = \cos(\frac{\pi}{3}) = \frac{1}{2}$, while the squared magnitude of each unit vector is $1$. Substituting these values into the expanded equation yields $3\lambda - \frac{1}{2}\lambda^2 + 3 - 2\lambda = 0$, which simplifies to $-\frac{1}{2}\lambda^2 + \lambda + 3 = 0$. Multiplying by $-2$ results in the quadratic equation $\lambda^2 - 2\lambda - 6 = 0$. Using the quadratic formula, the roots are $\lambda = 1 \pm \sqrt{7}$. Because $1 + \sqrt{7} \approx 3.65$ and $1 - \sqrt{7} \approx -1.65$, neither of these values falls within the range $[-1, 3]$, meaning there are no valid values for $\lambda$.

The vector $\vec{b}$ is determined by evaluating the sequence of nested cross products provided in the problem. Computing the first cross product $\vec{a} \times (\hat{i} + \hat{j})$ results in $\hat{i} - \hat{j} + \hat{k}$. Using this vector to calculate the next cross product with $\hat{i}$ yields $\hat{j} + \hat{k}$, and finally, the subsequent cross product with $\hat{i}$ gives $\vec{b} = \hat{j} - \hat{k}$.

With $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{j} - \hat{k}$ identified, we calculate the scalar projection of $\vec{a}$ onto $\vec{b}$ using the formula $P = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. The dot product $\vec{a} \cdot \vec{b}$ is evaluated as $(2)(0) + (1)(1) + (-1)(-1) = 2$, and the magnitude $|\vec{b}|$ is $\sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$. Therefore, the projection equals $\frac{2}{\sqrt{2}} = \sqrt{2}$, and its square is 2.

\;\overline{ {\text{N}}}={|\overline{ {\text{A}}}|}\ { {\text{B}}^{\^}}=\;{5(4 { {\text{i}}^{\^}}+3 { {\text{j}}^{\^}})}/{5}=4 { {\text{i}}^{\^}}+3 { {\text{j}}^{\^}} $\; \therefore {\text{x}}=4$

A unit vector $\vec{r}$ oriented along a resultant vector $\vec{v}$ is calculated by dividing that vector by its magnitude, written as $\vec{r} = \frac{\vec{v}}{|\vec{v}|}$. Given the direction is determined by the sum $\vec{b} + \vec{c}$, we first compute this resultant vector as $5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}$. Since the dot product of this unit vector with $\vec{a}$ is 3, we establish the relationship $\frac{(\vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{b} + \vec{c}|} = 3$.

The numerator involves finding the scalar product $(\vec{b} + \vec{c}) \cdot \vec{a} = (5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k})$, which simplifies to $3\lambda - 6$. Simultaneously, the magnitude of the resultant vector in the denominator is $\sqrt{5^2 + 2^2 + (\lambda - 5)^2}$, which reduces to $\sqrt{\lambda^2 - 10\lambda + 54}$. Equating the ratio $\frac{3\lambda - 6}{\sqrt{\lambda^2 - 10\lambda + 54}} = 3$ allows us to simplify by dividing by 3 on both sides, resulting in $\frac{\lambda - 2}{\sqrt{\lambda^2 - 10\lambda + 54}} = 1$. Squaring both sides yields the equation $\lambda^2 - 4\lambda + 4 = \lambda^2 - 10\lambda + 54$. By rearranging the variables and constants, we arrive at $6\lambda = 50$, which confirms that $3\lambda = 25$.

The relationship between the angle $\alpha$ and the vectors $\vec{b}$ and $\vec{c}$ is governed by the dot product formula, which states that $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}| \cos\alpha$. Substituting the definition $\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$ into this dot product yields $\vec{b} \cdot \vec{c} = 2\vec{b} \cdot (\vec{a} \times \vec{b}) - 3|\vec{b}|^2$. Because the scalar triple product $\vec{b} \cdot (\vec{a} \times \vec{b})$ is zero due to the presence of two identical vectors in the cross product, the expression simplifies to $\vec{b} \cdot \vec{c} = -3|\vec{b}|^2$. Given $|\vec{b}|=4$, this becomes $\vec{b} \cdot \vec{c} = -3(16)$, which is -48, leading to the equation $4|\vec{c}|\cos\alpha = -48$, or $|\vec{c}|\cos\alpha = -12$.

Squaring this relation yields $|\vec{c}|^2 \cos^2\alpha = 144$. The squared magnitude $|\vec{c}|^2$ is obtained by expanding $|2(\vec{a} \times \vec{b}) - 3\vec{b}|^2$, which results in $4|\vec{a} \times \vec{b}|^2 + 9|\vec{b}|^2$. The magnitude squared of the cross product is determined by the identity $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$, giving $1^2(4)^2 - 2^2$, which equals 12. Substituting these values into the expression for $|\vec{c}|^2$ produces $4(12) + 9(16) = 48 + 144 = 192$. With $|\vec{c}|^2 = 192$, the equation $192 \cos^2\alpha = 144$ simplifies to $\cos^2\alpha = 3/4$. Applying the trigonometric identity $\sin^2\alpha = 1 - \cos^2\alpha$ results in $192 \sin^2\alpha = 192(1 - 3/4)$, which equals 48.

The area of a parallelogram defined by its diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula $\frac{1}{2}|\vec{d_1} \times \vec{d_2}|$. Combining the vectors provided in the problem, the diagonals are $\vec{d_1} = \hat{i} + \alpha\hat{j} + 2\hat{k}$ and $\vec{d_2} = -\hat{i} + \beta\hat{j}$. Calculating the cross product leads to the determinant
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 2 \\ -1 & \beta & 0 \end{vmatrix} = -2\beta\hat{i} - 2\hat{j} + (\alpha+\beta)\hat{k}$
The magnitude of this cross product, set equal to the given area, results in the equation $\frac{1}{2}\sqrt{(-2\beta)^2 + (-2)^2 + (\alpha+\beta)^2} = \frac{\sqrt{21}}{2}$, which simplifies to $4\beta^2 + 4 + (\alpha+\beta)^2 = 21$.

Expanding this expression gives $4\beta^2 + 4 + \alpha^2 + 2\alpha\beta + \beta^2 = 21$. Substituting the known product $\alpha\beta = -6$ into the equation yields $\alpha^2 + 5\beta^2 + 4 - 12 = 21$, which further simplifies to $\alpha^2 + 5\beta^2 = 29$. Given that $\alpha$ and $\beta$ must be integers, the only pairs $(\alpha, \beta)$ that satisfy $\alpha\beta = -6$ and $\alpha^2 + 5\beta^2 = 29$ are $(-3, 2)$ and $(3, -2)$. Evaluating the expression $\alpha_1^2 + \beta_1^2 - \alpha_2\beta_2$ with these pairs results in $9 + 4 - (-6)$, which equals 19.

The vector relationship $\vec{a} = \vec{b} \times \vec{c}$ implies that $\vec{a}$ is orthogonal to both $\vec{b}$ and $\vec{c}$, which ensures that the dot product $\vec{a} \cdot \vec{c}$ is zero. When we expand the squared magnitude $|\vec{c} - \vec{a}|^2$, we get $|\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{a} \cdot \vec{c})$, which simplifies directly to $|\vec{c}|^2 + 4$ since $|\vec{a}| = 2$. To minimize the full expression $27(|\vec{c}|^2 + 4)$, we must first determine the minimum magnitude of $\vec{c}$ based on the given constraint. Using the definition of the cross product magnitude, $|\vec{a}| = |\vec{b}| |\vec{c}| \sin \alpha$, we substitute the known values to get $2 = 3 |\vec{c}| \sin \alpha$, which rearranges to $|\vec{c}| = \frac{2}{3} \csc \alpha$.

Given that $\alpha$ is restricted to the interval $[0, \frac{\pi}{3}]$, the sine function increases over this range, meaning the cosecant function $\csc \alpha$ decreases and reaches its minimum at $\alpha = \frac{\pi}{3}$. At this value, $\csc(\frac{\pi}{3}) = \frac{2}{\sqrt{3}}$, establishing the minimum magnitude as $|\vec{c}| = \frac{2}{3} \times \frac{2}{\sqrt{3}} = \frac{4}{3\sqrt{3}}$. Squaring this result gives $|\vec{c}|^2 = \frac{16}{27}$. Substituting this into the target expression $27(|\vec{c}|^2 + 4)$ results in $27(\frac{16}{27} + 4)$, which simplifies to $16 + 108$, yielding a final value of $124$.

The equation $\vec{a} \times \vec{r} = \vec{a} \times \vec{b}$ implies that the vector $\vec{r} - \vec{b}$ is parallel to $\vec{a}$, allowing us to express it as $\vec{r} - \vec{b} = k\vec{a}$. By substituting this relation into the dot product constraint $\vec{a} \cdot \vec{r} = 0$, we obtain:
$\vec{a} \cdot (\vec{b} + k\vec{a}) = \vec{a} \cdot \vec{b} + k|\vec{a}|^2 = 0$
Using the given vectors, we calculate $\vec{a} \cdot \vec{b} = 7$ and $|\vec{a}|^2 = 14$, which leads to the scalar value $k = -1/2$. Consequently, $\vec{r} = \vec{b} - \frac{1}{2}\vec{a} = \frac{3}{2}\hat{i} + \frac{1}{2}\hat{k}$, and since the magnitude of this resulting vector is $\frac{\sqrt{10}}{2}$, the claim in Statement-I is incorrect.

Regarding Statement-II, the identity $\cos 2A + \cos 2B + \cos 2C = -1 - 4 \cos A \cos B \cos C$ holds for any triangle $ABC$. Because the product of the cosines of the angles in a triangle is at most $\frac{1}{8}$, the expression is bounded below by $-1 - 4(\frac{1}{8}) = -\frac{3}{2}$, which confirms that Statement-II is correct.

Vector cross products obey the property that if $\vec{u} \times \vec{c} = \vec{c} \times \vec{v}$, then the expression can be rearranged to $(\vec{u} + \vec{v}) \times \vec{c} = 0$, which implies that $\vec{c}$ is collinear with the vector sum $(\vec{u} + \vec{v})$. Applying this principle to the equation $(\vec{a}+\vec{b}) \times \vec{c} = \vec{c} \times (-2\vec{a}+3\vec{b})$, we move the terms to one side to get $((\vec{a}+\vec{b}) + (-2\vec{a}+3\vec{b})) \times \vec{c} = 0$, simplifying to $(- \vec{a} + 4\vec{b}) \times \vec{c} = 0$.

With the given vectors $\vec{a}=4\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=11\hat{i}-\hat{j}+\hat{k}$, the vector $\vec{c}$ must be a scalar multiple $\lambda$ of $(4\vec{b} - \vec{a})$. Substituting the values, we find $\vec{c} = \lambda(4(11\hat{i}-\hat{j}+\hat{k}) - (4\hat{i}-\hat{j}+\hat{k})) = \lambda(40\hat{i}-3\hat{j}+3\hat{k})$. To determine the value of $\lambda$, we use the constraint $(2\vec{a}+3\vec{b}) \cdot \vec{c} = 1670$. First, calculating $2\vec{a}+3\vec{b}$ gives $2(4\hat{i}-\hat{j}+\hat{k}) + 3(11\hat{i}-\hat{j}+\hat{k}) = 41\hat{i}-5\hat{j}+5\hat{k}$.

Taking the dot product of $41\hat{i}-5\hat{j}+5\hat{k}$ with $\lambda(40\hat{i}-3\hat{j}+3\hat{k})$ results in $\lambda(41 \times 40 + (-5) \times (-3) + 5 \times 3) = 1670$, which simplifies to $\lambda(1640 + 15 + 15) = 1670$. This confirms that $\lambda = 1$, meaning $\vec{c} = 40\hat{i}-3\hat{j}+3\hat{k}$. The squared magnitude is then $40^2 + (-3)^2 + 3^2 = 1600 + 9 + 9 = 1618$.

The magnitude of resultant vector$R^{'}={\sqrt{a^{2+}b^{2+}2 a b \cos \theta }}$Here $a=b=R$Then $R^{'}={\sqrt{R^{2+}R^{2+}2 R^2 \cos \theta }}$ $\;=R {\sqrt{2}} {\sqrt{1+\cos \theta }}$ $\;={\sqrt{2}} R \sqrt{2 \cos ^2 \;\frac{\theta }{2}}$ $\;=2 R \cos \;\frac{\theta }{2}$

$\;\vec{R}= (A+\;{A}/{{\sqrt{2}}}) \hat{i}+ (A-\;{A}/{{\sqrt{2}}}) \hat{j}$ $\;|\vec{R}|=\sqrt{ (A+\;{A}/{{\sqrt{2}}}) ^{2+} (A-\;{A}/{{\sqrt{2}}}) ^2}={\sqrt{3}} A$

$\;B \cos \theta =\;\frac{B}{2}$ $\;\Rightarrow \theta =60^{\circ}$So, angle between ${ {\text{A}}}^{\to }$ & ${ {\text{B}}}^{\to }$ is $90^{\circ}+60^{\circ}=150^{\circ}$

When two vectors produce a cross product of zero, it signifies that the vectors are parallel, meaning one must be a scalar multiple of the other. The original relationship $(\vec{a} + 2\vec{b}) \times \vec{c} = 3(\vec{c} \times \vec{a})$ can be manipulated by reversing the cross product on the right side using the identity $\vec{c} \times \vec{a} = -(\vec{a} \times \vec{c})$. This adjustment transforms the expression into $(\vec{a} + 2\vec{b}) \times \vec{c} = -3(\vec{a} \times \vec{c})$, which simplifies to:
$(4\vec{a} + 2\vec{b}) \times \vec{c} = 0$
This implies that $\vec{c}$ is parallel to the sum $4\vec{a} + 2\vec{b}$.

Calculating $4\vec{a} + 2\vec{b}$ involves taking $4(\hat{i} - 3\hat{j} + 7\hat{k}) + 2(2\hat{i} - \hat{j} + \hat{k})$, which simplifies to the vector $8\hat{i} - 14\hat{j} + 30\hat{k}$. Because $\vec{c}$ must be a scalar multiple of this vector, we can write $\vec{c} = \lambda(8\hat{i} - 14\hat{j} + 30\hat{k})$. By applying the condition $\vec{a} \cdot \vec{c} = 130$, we substitute our expression for $\vec{c}$ into the dot product:
$(\hat{i} - 3\hat{j} + 7\hat{k}) \cdot \lambda(8\hat{i} - 14\hat{j} + 30\hat{k}) = 130$
Expanding this yields $\lambda(8 + 42 + 210) = 130$, which simplifies to $260\lambda = 130$, leading to $\lambda = \frac{1}{2}$. Using this scalar, $\vec{c}$ is defined as $4\hat{i} - 7\hat{j} + 15\hat{k}$. The final value for $\vec{b} \cdot \vec{c}$ is found by computing $(2\hat{i} - \hat{j} + \hat{k}) \cdot (4\hat{i} - 7\hat{j} + 15\hat{k})$, which results in $8 + 7 + 15$, totaling 30.

A vector $\vec{C} = C_1\hat{i} + C_2\hat{j} + C_3\hat{k}$ constrained by specific angles to other vectors allows for the determination of its components using the dot product formula $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta$. Applying this to the angle of $60^{\circ}$ with $2\hat{i}+2\hat{j}-\hat{k}$ results in the equation $2C_1+2C_2-C_3=\frac{3}{2}$, while the angle of $45^{\circ}$ with $\hat{i}-\hat{k}$ leads to $C_1-C_3=1$. Given that $\vec{C}$ is a unit vector, its components must satisfy $C_1^{2+}C_2^{2+}C_3^2=1$, which yields the values $C_1=\frac{\sqrt{2}}{3}+\frac{1}{2}$, $C_2=\frac{-1}{3\sqrt{2}}$, and $C_3=\frac{\sqrt{2}}{3}-\frac{1}{2}$. Substituting these values into the expression $\vec{C}+\left(-\frac{1}{2}\hat{i}+\frac{1}{3\sqrt{2}}\hat{j}-\frac{\sqrt{2}}{3}\hat{k}\right)$ involves adding the corresponding components, specifically $\left(\frac{\sqrt{2}}{3}+\frac{1}{2}-\frac{1}{2}\right)\hat{i} + \left(-\frac{1}{3\sqrt{2}}+\frac{1}{3\sqrt{2}}\right)\hat{j} + \left(\frac{\sqrt{2}}{3}-\frac{1}{2}-\frac{\sqrt{2}}{3}\right)\hat{k}$, which simplifies to $\frac{\sqrt{2}}{3}\hat{i}-\frac{1}{2}\hat{k}$.

To approach this problem, we apply the vector triple product identity, $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$, which serves as the key to simplifying the complex cross-product expression. Starting with the given equation $\vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k}$, we first distribute the cross product to obtain $\vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k}$. By taking the cross product of both sides with $\vec{a}$, we generate terms of the form $\vec{a} \times (\vec{a} \times \vec{b})$, $\vec{a} \times (\vec{a} \times \vec{c})$, and $\vec{a} \times (\vec{b} \times \vec{c})$, which we then expand using the triple product identity.

Substituting the known values where $\vec{a} \cdot \vec{b} = -26$, $\vec{a} \cdot \vec{c} = 13$, and $|\vec{a}|^2 = 29$, the equation simplifies into a linear relationship involving $\vec{a}$, $\vec{b}$, and $\vec{c}$. Specifically, this reduction yields $-13\vec{a} - 16\vec{b} - 3\vec{c} = \vec{a} \times (\hat{i} + 8\hat{j} + 13\hat{k})$. To isolate the required term $\vec{b} \cdot \vec{c}$, we compute the dot product of this entire equation with $\vec{b}$, which converts the cross product term on the right into a scalar triple product. Upon calculating the scalar components with $|\vec{b}|^2 = 50$, the equation simplifies to $-462 - 3\vec{b} \cdot \vec{c} = -396$. Solving for $\vec{b} \cdot \vec{c}$ gives $-22$, and finally calculating $24 - (-22)$ leads to the result of 46.

When two vectors are inclined at an obtuse angle, their dot product must be strictly negative. For the given vectors $\vec{a} = \alpha t \hat{i} + 6 \hat{j} - 3 \hat{k}$ and $\vec{b} = t \hat{i} - 2 \hat{j} - 2 \alpha t \hat{k}$, computing the dot product results in the quadratic expression $\alpha t^2 + 6 \alpha t - 12 < 0$. For a quadratic expression to remain negative for all real values of $t$, the coefficient of $t^2$ must be negative to ensure the parabola opens downwards, and the discriminant must be negative so the graph never crosses the horizontal axis.

Applying the condition for the leading coefficient, we require $\alpha < 0$. Setting the discriminant $D = (6 \alpha)^2 - 4(\alpha)(-12)$ to be less than zero leads to the inequality $36 \alpha^2 + 48 \alpha < 0$, which simplifies to $12 \alpha (3 \alpha + 4) < 0$. Solving this quadratic inequality reveals that $\alpha$ must lie between the roots $-\frac{4}{3}$ and $0$. Testing the boundary case where $\alpha$ is $0$, the original dot product expression simplifies to the constant -12, which is always negative regardless of the value of $t$. Consequently, the complete set of valid values for $\alpha$ is $\left( -\frac{4}{3}, 0 \right]$.

The scalar triple product of three vectors corresponds to the volume of a parallelepiped and can be efficiently computed as the determinant of a matrix formed by these vectors. Since $\vec{d}$ represents the unit vector in the direction of $\vec{b} + \vec{c}$, it is defined as $\vec{d} = \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|}$. Given the condition $\vec{a} \cdot \vec{d} = 1$, substituting this expression leads to the relation $\vec{a} \cdot (\vec{b} + \vec{c}) = |\vec{b} + \vec{c}|$.

Summing the vectors $\vec{b}$ and $\vec{c}$ results in $(x+2)\hat{i} + 6\hat{j} - 2\hat{k}$. Computing the dot product of $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ with this sum gives $x + 2 + 6 - 2 = x + 6$. Equating this to the magnitude $|\vec{b} + \vec{c}| = \sqrt{(x+2)^2 + 6^2 + (-2)^2}$ yields the equation $x + 6 = \sqrt{x^2 + 4x + 44}$. Squaring both sides provides $(x+6)^2 = x^2 + 4x + 44$, which simplifies to $x^2 + 12x + 36 = x^2 + 4x + 44$. Solving this for $x$ results in $8x = 8$, confirming that $x = 1$.