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$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.2

Related Topics

Wize University Linear Algebra Textbook > Products of Vectors

Projection (Proj) and Perpendicular (Perp)

6 Activities

Wize University Linear Algebra Textbook > Equations of Lines and Planes

Hyperplanes

3 Activities

Given the points P1=(0,0,1)\bcb{P_1 = (0,0,1)} and P2=(1,−2,3)\bcb{P_2 = (1,-2,3)} and the vector v⃗=<2,−3,1>\bcb{\vec{v} = \left< 2, -3, 1 \right>}
Find the minimum distance between the point P2\bcb{P_2} and the plane Π: 2x−3y+z=1\bcb{\Pi:\ 2x-3y + z = 1}.
More Projection (Proj) and Perpendicular (Perp) Questions:
Example: Dot Product (2)
Find the scalar and vector projection of b⃗=⟨2,1,3⟩\vec{b}=\langle 2,1,3\rangleonto a⃗=⟨−1,4,2⟩\vec{a}=\langle -1,4,2\rangle.
Find the vector projection of [3,2][3,2] onto the vector [3,6]\left[3,6\right]
133 - FML 3 - 18.1W - e.g. 9
Find the vector-component ı^∣∣d⃗ \ihat_{||_{\vd}} \, of ı^\bcb{\boldsymbol{ \ihat}} in the direction of the line given by y=mx\bcb{\boldsymbol{ y=mx}}. Repeat for ȷ^∣∣d⃗ \jhat_{||{\vd}} \, , the vector-component of ȷ^\bcb{\boldsymbol{ \jhat }} in the direction of the line.
133 - FML 3 - 18.1W e.g. 58.3
Given the vector v⃗=<1,1,2>\bcb{\vec{v} = \left< 1, 1, 2 \right>} and the points P1=(0,−1,0)\bcb{P_1 = (0,-1,0)} and P2=(−1,1,0)\bcb{P_2 = (-1,1,0)}
Find the point on the line L\bcb{L} closest to the point P2\bcb{P_2}.
133 - FML 3 - 18.1W e.g. 58.2
Given the vector v⃗=<1,1,2>\bcb{\vec{v} = \left< 1, 1, 2 \right>} and the points P1=(0,−1,0)\bcb{P_1 = (0,-1,0)} and P2=(−1,1,0)\bcb{P_2 = (-1,1,0)}
Find the minimum distance of P2\bcb{P_2} to the line L\bcb{L} found above.
133 - FML 3 - 18.1W e.g. 48
Find a unit vector in the direction of the projection of u⃗=<102,112,−56>\bcb{\vec{u} = \left< 102, 112, -56 \right>} onto v⃗=<0,3,4>\bcb{\vec{v} = \left< 0, 3, 4 \right>}.
133 - FML 3 - 18.1W e.g. 43
Given u⃗=<8,2,2>\bcb{\vec{u} = \left< 8, 2, 2 \right>} and v⃗=<2,−1,−1>\bcb{\vec{v} = \left< 2, -1, -1 \right>}, find the vector projection of u⃗\bcb{\vec{u}} onto v⃗\bcb{\vec{v}}.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 41
Let P=(−3,−3,−4)\bcb{P = (-3, -3, -4)}, Q=(−5,−6,−3)\bcb{Q = (-5,-6,-3)}, R=(−6,−4,−6)\bcb{R = (-6,-4,-6)} and S=(−2,−3,−4)\bcb{S = (-2, -3, -4)}. Let l1\bcb{l_1} be the line passing through P\bcb{P} and Q\bcb{Q}, and let l2\bcb{l_2} be the line passing through R\bcb{R} and S\bcb{S}. Find the distance between R\bcb{R} and l1\bcb{l_1}.
133 - FML 3 - 18.1W e.g. 36
For two lines having directional vectors v1\mathbf{v}_1 and v2\mathbf{v}_2, respectively, the shortest distance between them can be found by creating a vector that connects the two lines (by subtracting a point on one line from a point on the other), i.e. (r2−r1)\left(\mathbf{r}_2 - \mathbf{r}_1\right), and projecting it onto the unit normal vector connecting the two lines. The unit normal is easily found by crossing their respective directionals and dividing by the magnitude of the resulting vector, as illustrated in Fig. 5. i.e.
s=∣projn^(r2−r1)∣=∣(r2−r1)⋅v1×v2∣v1×v2∣∣ s = \left| \text{proj}_{\mathbf{\hat{n}}} \left(\mathbf{r}_2 - \mathbf{r}_1\right)\right| = \left| \left(\mathbf{r}_2 - \mathbf{r}_1\right) \cdot \frac{\mathbf{v}_1 \times \mathbf{v}_2}{\left|\mathbf{v}_1 \times \mathbf{v}_2\right|}\right|
Find the equations of the line passing through the point (2,3,−4)\bcb{(2, 3, -4)} parallel to the vector i−4k\bcb{\mathbf{i} - 4\mathbf{k}} ; express it in vector-parametric, scalar-parametric and standard form.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.3
Given the points P1=(0,0,1)\bcb{P_1 = (0,0,1)} and P2=(1,−2,3)\bcb{P_2 = (1,-2,3)} and the vector v⃗=<2,−3,1>\bcb{\vec{v} = \left< 2, -3, 1 \right>}, find the point Po\bcb{P_o} on the plane Π: 2x−3y+z=1\bcb{\Pi: \ 2x - 3y + z = 1} that is closest to the point P2\bcb{P_2}.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 48
Find a unit vector in the direction of the projection of u⃗=<102,112,−56>\bcb{\vec{u} = \left< 102, 112, -56 \right>} onto v⃗=<0,3,4>\bcb{\vec{v} = \left< 0, 3, 4 \right>}.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 43
Given u⃗=<8,2,2>\bcb{\vec{u} = \left< 8, 2, 2 \right>} and v⃗=<2,−1,−1>\bcb{\vec{v} = \left< 2, -1, -1 \right>}, find the vector projection of u⃗\bcb{\vec{u}} onto v⃗\bcb{\vec{v}}.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.2
Given the vector v⃗=<1,1,2>\bcb{\vec{v} = \left< 1, 1, 2 \right>} and the points P1=(0,−1,0)\bcb{P_1 = (0,-1,0)} and P2=(−1,1,0)\bcb{P_2 = (-1,1,0)}, find the minimum distance of P2\bcb{P_2} to the line L\bcb{L} that is parallel to v⃗\bcb{\vec{v}} that passes through P1\bcb{P_1}.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.3
Given the vector v⃗=<1,1,2>\bcb{\vec{v} = \left< 1, 1, 2 \right>} and the points P1=(0,−1,0)\bcb{P_1 = (0,-1,0)} and P2=(−1,1,0)\bcb{P_2 = (-1,1,0)}, find the point on
the line L\bcb{L} (which is parallel to v⃗\bcb{\vec{v}} that passes through P1\bcb{P_1}) that is closest to the point P2\bcb{P_2}.
More Hyperplanes Questions:
133 - FML 3 - 18.1W e.g. 36
For two lines having directional vectors v1\mathbf{v}_1 and v2\mathbf{v}_2, respectively, the shortest distance between them can be found by creating a vector that connects the two lines (by subtracting a point on one line from a point on the other), i.e. (r2−r1)\left(\mathbf{r}_2 - \mathbf{r}_1\right), and projecting it onto the unit normal vector connecting the two lines. The unit normal is easily found by crossing their respective directionals and dividing by the magnitude of the resulting vector, as illustrated in Fig. 5. i.e.
s=∣projn^(r2−r1)∣=∣(r2−r1)⋅v1×v2∣v1×v2∣∣ s = \left| \text{proj}_{\mathbf{\hat{n}}} \left(\mathbf{r}_2 - \mathbf{r}_1\right)\right| = \left| \left(\mathbf{r}_2 - \mathbf{r}_1\right) \cdot \frac{\mathbf{v}_1 \times \mathbf{v}_2}{\left|\mathbf{v}_1 \times \mathbf{v}_2\right|}\right|
Find the equations of the line passing through the point (2,3,−4)\bcb{(2, 3, -4)} parallel to the vector i−4k\bcb{\mathbf{i} - 4\mathbf{k}} ; express it in vector-parametric, scalar-parametric and standard form.