What value of \(k\), where \(k \in R\), will make the following planes perpendicular?
\(\Pi_1: \ 2 x-k y+3 z=1\)
\(\Pi_2: \ 2 k x+3 y-2 z=4\)
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What value of \(k\), where \(k \in R\), will make the following planes perpendicular?
\(\Pi_1: \ 2 x-k y+3 z=1\)
\(\Pi_2: \ 2 k x+3 y-2 z=4\)
\(C\)
\(\underset{\sim}{p_1} = 2\underset{\sim}{i}-k\underset{\sim}{j}+3\underset{\sim}{k}\)
\(\underset{\sim}{p_2} = 2k\underset{\sim}{i}+3\underset{\sim}{j}-2\underset{\sim}{k}\)
\(\text{Solve}\ \ \underset{\sim}{p_1} \cdot \underset{\sim}{p_2}=0\ \ \text{for}\ k\ \text{(by calc)}: \)
\(4k-3k-6=0\ \ \Rightarrow \ k=6\)
\(\Rightarrow C\)
Let \(\underset{\sim}{\text{a}}=\underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}, \underset{\sim}{\text{b}}=\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}\) and \(\text{c}=\underset{\sim}{\text{i}}+2 \underset{\sim}{\text{j}}+3 \underset{\sim}{\text{k}}\).
If \(\underset{\sim}{\text{n}}\) is a unit vector such that \( \underset{\sim} {\text{a}} \cdot \underset{\sim} {\text{n}}=0\) and \( \underset{\sim} {\text{b}} \cdot \underset{\sim} {\text{n}}=0\), then \(\big{|} \underset{\sim} {\text{c}} \cdot \underset{\sim} {\text{n}}\big{|} \) is equal to
\(B\)
\(\underset{\sim}{\text{n}}\ \text{is perpendicular to}\ \underset{\sim}{\text{a}}\ \text{and}\ \underset{\sim}{\text{b}} .\)
\(\underset{\sim}{\text{a}}\ \text{and}\ \underset{\sim}{\text{b}}\ \text{both lie on the}\ \underset{\sim}{\text{i}}\ –\ \underset{\sim}{\text{j}}\ \text{plane.}\)
\(\text{Possible (unit vector)}\ \underset{\sim}{\text{n}} = (0,0,1) \)
\(\therefore \big{|} \underset{\sim} {\text{c}} \cdot \underset{\sim} {\text{n}}\big{|} = 3\)
\(\Rightarrow B\)
Let `underset ~ a = 2 underset ~i - 3 underset ~j + underset ~k` and `underset ~b = underset ~i + m underset ~j - underset ~k`, where `m` is an integer.
The vector resolute of `underset ~a` in the direction of `underset ~b` is `-11/18 (underset ~i + m underset ~j - underset ~k)`.
a. | `underset ~b ⋅ underset ~a` | `= ((1), (m), (-1))((2), (-3), (1)) = 2 – 3m – 1 = 1 – 3m` |
`|underset ~b|^2` | `= 1^2 + m^2 + (-1)^2 = m^2 + 2` |
`(underset ~b ⋅ underset ~a)/|underset ~b|^2 ⋅ underset ~b` | `= -11/18 underset ~b` |
`(1 -3m)/(m^2 + 2)` | `= -11/18` |
`18 – 54m` | `= -11m^2 – 22` |
`11m^2 – 54m + 40` | `=0` | |
`(11m – 10)(m – 4)` | `=0` |
`:. m= 4\ \ (m != 10/11, \ m ∈ Z)`
b. | `underset ~a_(⊥ underset ~b)` | `= underset ~a + 11/18 (underset ~i + 4 underset ~j – underset ~k)` |
`= 2 underset ~i + 11/18 underset ~i – 3 underset ~j + 44/18 underset ~j + underset ~k – 11/18 underset ~k` | ||
`= 47/18 underset ~i – 5/9 underset ~j + 7/18 underset ~k` |
The vectors `underset~a = 2underset~i + m underset~j - 3underset~k` and `underset~b = m^2underset~i - underset~j + underset~k` are perpendicular for
`D`
`underset ~a ⊥ underset ~b\ \ =>\ \ underset ~a ⋅ underset ~b=0`
`underset ~a ⋅ underset ~b` | `= 2m^2 + m(-1) + (-3)(1)` |
`0` | `= 2m^2 – m – 3` |
`0` | `= (2m – 3)(m + 1)` |
`:. m = 3/2, quad m = -1`
`=> D`
Consider the three vectors
`underset ~a = underset ~i - underset ~j + 2underset ~k,\ underset ~b = underset ~i + 2 underset ~j + m underset ~k` and `underset ~c = underset ~i + underset ~j - underset ~k`, where `m in R.`
a. | `|underset~b|` | `= sqrt(1^2 + 2^2 + m^2)` | `= 2sqrt3` |
`1 + 4 + m^2` | `= 4 xx 3` | ||
`m^2` | `= 7` | ||
`m` | `= ±sqrt7` |
b. | `underset~a * underset~b` | `= 1 xx 1 + (−1) xx 2 + 2 xx m` | |
`0` | `= 1 – 2 + 2m\ \ ( underset~a ⊥ underset~b)` | ||
`2m` | `= 1` | ||
`m` | `= 1/2` |
c.i. `3 underset ~c – underset ~a`
`=3underset~i – underset~i + 3underset~j + underset~j -3underset~k – 2underset~k`
`=2underset~i + 4underset~j-5underset~k`
c.ii. `text(Linear dependence)\ \ =>\ \ 3underset~c – underset~a = t underset~b,\ \ t in RR`
`2 tilde i + 4 tilde j – 5 tilde k= t (tilde i + 2 tilde j + m tilde k)`
`=> t = 2 and tm = -5,`
`:. m=-5/2`
Two vectors are given by `underset ~a = 4 underset ~i + m underset ~j - 3 underset ~k` and `underset ~b = −2 underset ~i + n underset ~j - underset ~k`, where `m`, `n in R^+`.
If `|\ underset ~a\ | = 10` and `underset ~a` is perpendicular to `underset ~b`, then `m` and `n` respectively are
`A`
`text(Using)\ \ |\ underset ~a\ | = 10:`
`10` | `= sqrt(4^2 + m^2 + (-3)^2)` |
`m` | `= 5sqrt3\ \ \ (text{by CAS,}\ \ m in R^+)` |
`text(S)text(ince)\ \ underset ~a _|_ underset ~b :`
`underset ~a ⋅ underset ~b` | `= 0` |
`0` | `=4 xx (−2) + mn + (−3) xx (−1)` |
`0` | `= mn-5` |
`n = sqrt 3/3\ \ \ (text{by CAS,}\ \ n in R^+)`
`=> A`
If `underset ~a = -2 underset ~i - underset ~j + 3 underset ~k` and `underset ~b = -m underset ~i + underset ~j + 2 underset ~k`, where `m` is a real constant, the vector `underset ~a - underset ~b` will be perpendicular to vector `underset ~b` where `m` equals
A. 0 only
B. 2 only
C. 0 or 2
D. 4.5
E. 0 or −2
`C`
`underset ~a – underset ~b` | `= (m – 2) underset ~i – 2 underset ~j + underset ~k ` |
`(underset ~a – underset ~b) ⋅ underset ~b` | `= -m(m – 2) – 2(1) + 1(2) = 0` |
`0` | `= -m^2 + 2m – 2 + 2` |
`0` | `= 2m – m^2` |
`0` | `= m(2 – m) \ => \ m = 0 \ or \ 2` |
`=> C`
Consider the vector `underset ~a = sqrt 3 underset ~i - underset ~j - sqrt 2 underset ~k`, where `underset ~i, underset ~j` and `underset ~k` are unit vectors in the positive directions of the `x, y` and `z` axes respectively.
Given that `underset ~b` is perpendicular to `underset ~a,` find the value of `m`. (2 marks)
a. | `|underset ~a|` | `= sqrt((sqrt 3)^2 + (-1)^2 + (-sqrt 2)^2)` |
`= sqrt 6` |
`:. hat underset ~a` | `= underset ~a/|underset ~a|` |
`= 1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)` |
b. | `underset ~a ⋅ underset ~i` | `= sqrt 3 xx 1 = sqrt 3` |
`underset ~a ⋅ underset ~i` | `= |underset ~a||underset ~i| cos theta` | |
`= sqrt 6 cos theta` | ||
`sqrt 3` | `= sqrt 6 cos theta` |
`cos theta` | `= 1/sqrt 2` |
`:. theta` | `= pi/4 = 45^@` |
c. `underset ~a ⋅ underset ~b = sqrt 3 (2 sqrt 3) + (-1)(m) + (-sqrt 2)(-5) = 0`
`6 – m + 5 sqrt 2` | `=0` | |
`:. m` | `=6 + 5 sqrt 2` |