"Vector Algebra" — Practice

Question 1

Find a unit vector in the direction of $2\hat i-3\hat j+6\hat k$.

Accepted Answer

The magnitude is the square root of $4+9+36$, which is $\sqrt{49}=7$. Dividing each component by 7 gives the unit vector $	frac27\hat i - 	frac37\hat j + 	frac67\hat k$. Check: the squares of the components are $4/49$, $9/49$ and $36/49$, summing to 1. The negative of this vector is also a unit vector but points the opposite way, so it is not an answer unless the question asks for a direction rather than the direction.

Question 2

Find the midpoint of $A(1,-2,3)$ and $B(3,4,-1)$ and verify it with the section formula.

Accepted Answer

The midpoint is the case $m=n$ of the section formula, so it is the average of the position vectors: $	frac12[(1,-2,3)+(3,4,-1)] = 	frac12(4,2,2) = (2,1,1)$. Verifying with the general formula at ratio $1:1$ gives $(1\cdot(3,4,-1) + 1\cdot(1,-2,3))/2$, the same result. As a further check, the vector from $A$ to the midpoint is $(1,3,-2)$ and from the midpoint to $B$ is also $(1,3,-2)$, equal as required.

Question 3

If $\vec a=\hat i+\hat j+\hat k$, $\vec b=2\hat i-\hat j+3\hat k$ and $\vec c=\hat i-2\hat j+\hat k$, find a unit vector parallel to $2\vec a-\vec b+3\vec c$.

Accepted Answer

Combine componentwise: $2\vec a = (2,2,2)$, $-\vec b = (-2,1,-3)$ and $3\vec c = (3,-6,3)$. Adding gives $(3,-3,2)$. Its magnitude is the square root of $9+9+4$, which is $\sqrt{22}$. The unit vector is therefore $(3\hat i-3\hat j+2\hat k)/\sqrt{22}$. Parallel allows either sign, so the negative is equally acceptable unless a direction is specified.

Question 4

Find the angle between $\hat i+\hat j$ and $\hat j+\hat k$.

Accepted Answer

The dot product is $1(0)+1(1)+0(1) = 1$. Each vector has magnitude $\sqrt2$, so the product of magnitudes is 2. Then $\cos	heta = 1/2$ and $	heta = 60^\circ$. This is worth remembering as a picture: the two vectors are face diagonals of a cube meeting at a vertex, and the triangle they form with the third face diagonal is equilateral, which is why the angle is exactly sixty degrees.

Question 5

Find $\lambda$ if the projection of $\lambda\hat i+\hat j+4\hat k$ on $2\hat i+6\hat j+3\hat k$ is 4 units.

Accepted Answer

The projection divides by the magnitude of the second vector, which is the square root of $4+36+9$, that is 7. The dot product is $2\lambda + 6 + 12 = 2\lambda + 18$. Setting $(2\lambda+18)/7 = 4$ gives $2\lambda + 18 = 28$, so $\lambda = 5$. Note that dividing by the magnitude of the first vector instead would have produced a quadratic and a quite different answer, which is why identifying the screen vector matters.

Question 6

If $\lvert\vec a\rvert=3$, $\lvert\vec b\rvert=4$ and $\lvert\vec a+\vec b\rvert=5$, find $\lvert\vec a-\vec b\rvert$ and the angle between the vectors.

Accepted Answer

Expand the square of the sum: $\lvert\vec a+\vec b\rvert^2 = 9 + 16 + 2\vec a\cdot\vec b = 25$, so $\vec a\cdot\vec b = 0$ and the vectors are perpendicular, at $90^\circ$. Then $\lvert\vec a-\vec b\rvert^2 = 9 + 16 - 0 = 25$, so the difference also has magnitude 5. The 3, 4, 5 triangle is the visible reason: the two vectors are the legs of a right angle and both diagonals of the resulting rectangle have length 5.

Question 7

Three vectors of equal magnitude sum to zero. Show that the angle between any two of them is $120^\circ$.

Accepted Answer

Let each have magnitude $k$. From $\vec a+\vec b+\vec c=\vec 0$ we get $\vec a+\vec b = -\vec c$, so their magnitudes are equal: $\lvert\vec a+\vec bvert^2 = k^2$. Expanding the left side gives $k^2 + k^2 + 2\vec a\cdot\vec b = k^2$, so $\vec a\cdot\vec b = -k^2/2$ and $\cos	heta = -1/2$, giving $	heta = 120^\circ$. The argument is symmetric in the three vectors, so the same holds for every pair. Geometrically the three are the sides of an equilateral triangle traversed in order.

Question 8

Find a unit vector perpendicular to both $\hat i+\hat j$ and $\hat j+\hat k$.

Accepted Answer

The cross product $(1,1,0)	imes(0,1,1)$ has $i$ component $1(1)-0(1)=1$, $j$ component $-[1(1)-0(0)]=-1$ and $k$ component $1(1)-1(0)=1$, giving $(1,-1,1)$ with magnitude $\sqrt3$. The unit vectors are therefore $\pm(\hat i-\hat j+\hat k)/\sqrt3$. Both signs are correct, since a plane has two unit normals and nothing in the question distinguishes them. Verify by dotting with each original vector: $1-1+0=0$ and $0-1+1=0$.

Question 9

Find the area of the parallelogram whose diagonals are $3\hat i+\hat j-2\hat k$ and $\hat i-3\hat j+4\hat k$.

Accepted Answer

When the diagonals are given rather than the sides, the area is half the magnitude of their cross product. That cross product has $i$ component $1(4)-(-2)(-3) = 4-6 = -2$, $j$ component $-[3(4)-(-2)(1)] = -(12+2) = -14$, and $k$ component $3(-3)-1(1) = -10$, giving $(-2,-14,-10)$. Its magnitude is the square root of $4+196+100 = 300$, which is $10\sqrt3$. The area is half of that, $5\sqrt3$.

Question 10

Verify $\lvert\vec a	imes\vec bvert^{2}=\lvert\vec avert^{2}\lvert\vec bvert^{2}-(\vec a\cdot\vec b)^{2}$ for $\vec a=2\hat i-\hat j+\hat k$ and $\vec b=\hat i+2\hat j-\hat k$.

Accepted Answer

The dot product is $2-2-1 = -1$, and both magnitudes squared are $4+1+1 = 6$. So the right-hand side is $36 - 1 = 35$. The cross product has $i$ component $(-1)(-1)-(1)(2) = -1$, $j$ component $-[(2)(-1)-(1)(1)] = 3$, and $k$ component $(2)(2)-(-1)(1) = 5$, giving $(-1,3,5)$ with squared magnitude $1+9+25 = 35$. The two agree, which is Lagrange's identity rearranged.

Question 11

Find $\lambda$ so that $\hat i+3\hat j$, $2\hat i-\hat j-\hat k$ and $\lambda\hat i+7\hat j+3\hat k$ are coplanar.

Accepted Answer

Coplanarity means the scalar triple product vanishes, so set the determinant of the three rows to zero. Expanding along the first row gives $1[(-1)(3)-(-1)(7)] - 3[(2)(3)-(-1)\lambda] + 0$, which is $1(-3+7) - 3(6+\lambda) = 4 - 18 - 3\lambda = -14 - 3\lambda$. Setting this to zero gives $\lambda = -14/3$. For every other value the three vectors span a genuine volume and are not coplanar.

Question 12

Find the volume of the tetrahedron with one vertex at the origin and the others at $(1,2,3)$, $(2,1,4)$ and $(3,4,1)$.

Accepted Answer

With one vertex at the origin, the three position vectors are already the edge vectors. The determinant is $1(1\cdot1 - 4\cdot4) - 2(2\cdot1 - 4\cdot3) + 3(2\cdot4 - 1\cdot3)$, that is $1(-15) - 2(-10) + 3(5) = -15 + 20 + 15 = 20$. That 20 is the volume of the parallelepiped. A tetrahedron is one sixth of it, so the volume is $20/6 = 10/3$. Reporting 20, or half of it, are the two standard errors here.

Practice — "Vector Algebra"

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