Vector Algebra: questions with answers

Line 1 is correct. Line 2 is the first wrong line: the vectors are equal only up to a scalar, $4\vec a + x\vec b = \lambda(2\vec a + 3\vec b)$, so $4 = \lambda2$ and $x = \lambda3 = 6$. Line 3's check would have exposed the missing $\lambda$.

Line 1 is correct. Line 2 is the first wrong line: $|\lambda|$ is smallest at $\lambda = 0$, which lies INSIDE $[-4, 3]$, so the least value of $|\lambda\vec a|$ is $0$, not an end-point value. The range is $[0, 8]$. Line 3 follows from the wrong Line 2.

$|\vec a| = \sqrt{ 4^2 + 2^2 } = |\vec b|$, so the magnitudes are equal. But equal vectors need equal corresponding components, and $4 \neq 2$, so $\vec a \neq \vec b$: they point in different directions. Equal magnitudes never by themselves make two vectors equal.

Lines 1 and 2 are correct. Line 3 is the first wrong line: equal magnitudes do not make equal vectors -- $\overrightarrow{AB}$ and $\overrightarrow{BC}$ have different components and different directions. Line 4 happens to be true, but it follows from Line 2, not from Line 3.

All vectors perpendicular to both $\vec a$ and $\vec b$ lie along the line of $\vec a \times \vec b$, and that line carries exactly two unit vectors, $\pm\dfrac{\vec a \times \vec b}{|\vec a \times \vec b|}$, pointing opposite ways. Two, not one, and not infinitely many.

Equal inclination means $l = m = n$, and $l^2 + m^2 + n^2 = 1$ gives $3l^2 = 1$, $l = \dfrac{1}{\sqrt3}$ (positive for acute angles). So the unit vector is $\dfrac{1}{\sqrt3}(\hat i + \hat j + \hat k)$ and the required vector is $\dfrac{ 5 }{\sqrt3}(\hat i + \hat j + \hat k)$. $l = \dfrac13$ would make the cosines add to $1$ instead of their squares, and $5(\hat i + \hat j + \hat k)$ has magnitude $5\sqrt3$.

The ratio $m : n$ is $AP : PB$, read from the first-named end: $AP : PB = 2 : 3$, so $AP$ is the shorter part and $P$ is nearer to $A$. Reading the ratio the other way round places $P$ on the wrong side of the midpoint.

Line 1 is correct. Line 2 is the first wrong line: $\vec b\cdot\vec a = \vec a\cdot\vec b$, so $-\vec a\cdot\vec b + \vec b\cdot\vec a = 0$ and the product is $|\vec a|^2 - |\vec b|^2$. Then $|\vec b|^2 = 7^2 - (-15)$ and $|\vec b| = 8$. Line 3 follows from the wrong Line 2.

$\vec a\cdot\vec b = \vec a\cdot\vec c$ gives $\vec a\cdot(\vec b - \vec c) = 0$, which means $\vec b - \vec c = \vec 0$ OR $\vec a \perp (\vec b - \vec c)$. The dot product has no cancellation law: $\hat i\cdot\hat j = \hat i\cdot\hat k = 0$ with $\hat j \neq \hat k$.

Collinear vectors are parallel to one line, so $\vec b = \lambda\vec a$ for some scalar $\lambda$ -- of either sign and any size. $\vec a = \pm\vec b$ would force equal magnitudes; the components of collinear vectors ARE proportional; and they may have opposite directions, so a positive $\lambda$ cannot be required.

$\hat a\cdot\hat b = |\hat a||\hat b|\cos\theta = \cos\theta$, and $-1 \leq \cos\theta \leq 1$, so $|\hat a\cdot\hat b| \leq 1$ always. It equals $1$ only when the vectors coincide, can be negative for an obtuse angle, and can never exceed $1$ in size.

A is true: reversing the order reverses the direction given by the right-hand rule ($\hat i \times \hat j = \hat k$ but $\hat j \times \hat i = -\hat k$). R is false: the vector product is anti-commutative, not commutative. So A is true and R is false.

The scalar projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot\vec b}{|\vec b|} = \dfrac{\alpha + 12}{3}$, since $|\vec b| = 3$. Setting this equal to $2$: $\alpha + 12 = 6$, so $\alpha = -6$. Dividing by $|\vec b|^2 = 9$ gives the coefficient of the projection VECTOR, not the scalar projection.

Writing the diagonals as $\vec a + \vec b$ and $\vec a - \vec b$ in terms of the adjacent sides, $\vec d_1 \times \vec d_2 = (\vec a + \vec b) \times (\vec a - \vec b) = -2\,\vec a \times \vec b$, so the area $|\vec a \times \vec b| = \tfrac12|\vec d_1 \times \vec d_2|$. Without the half one gets twice the area; $|\vec d_1||\vec d_2|$ ignores the angle, and the dot product is not an area.

Line 1 is true. Line 2 is the first wrong line: the triangle law reads $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ -- head to tail -- so $\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB} = (2, -2)$. Two vectors both starting at $A$ add to a diagonal, not to a side. Line 3 evaluates the wrong Line 2 correctly.