A point p- q- r lies on the plane r-2-k-4- The value of q such that the vector a- p- q- rk- satisfies the relation -a-0- is

Plane is x + 2y + z = 4 As (p, q, r) lies on above plane, so p + 2q + r = 4 Now, ĵ×(ĵ×a)=0 ⇒(ĵ·a) ĵ (ĵ·ĵ)a=0 ⇒(ĵ·(pî+qĵ+rk̂))ĵ (pî+qĵ+rk̂)=0 ⇒ pî+rk̂= nbsp; ...

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Standard XII

Mathematics

Test for Collinearity of Points

A point p, q,...

Question

A point $(p, q, r)$ lies on the plane $\to r \cdot (^i + 2^j +^k) = 4$ . The value of $q$ such that the vector $\to a = p^i + q^j + r^k$ satisfies the relation $^j \times (^j \times \to a) = \to 0,$ is

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Solution

Plane is $x + 2 y + z = 4$
As $(p, q, r)$ lies on above plane,
so $p + 2 q + r = 4$
Now, $^j \times (^j \times \to a) = \to 0$
$\Rightarrow (^j \cdot \to a)^j - (^j \cdot^j) \to a = \to 0 \Rightarrow (^j \cdot (p^i + q^j + r^k))^j - (p^i + q^j + r^k) = \to 0$
$\Rightarrow p^i + r^k = \to 0,$
which is possible only when $p = r = 0$
So, from equation $(1)$ , we get
$2 q = 4 \Rightarrow q = 2$

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A point (p,q,r) lies on the plane →r⋅(^i+2^j+^k)=4. The value of q such that the vector →a=p^i+q^j+r^k satisfies the relation ^j×(^j×→a)=→0, is

Plane is x+2y+z=4 As (p,q,r) lies on above plane, so p+2q+r=4 Now, ^j×(^j×→a)=→0 ⇒(^j⋅→a)^j−(^j⋅^j)→a=→0⇒(^j⋅(p^i+q^j+r^k))^j−(p^i+q^j+r^k)=→0 ⇒p^i+r^k=→0, which is possible only when p=r=0 So, from equation (1), we get 2q=4⇒q=2

A point $(p, q, r)$ lies on the plane $\to r \cdot (^i + 2^j +^k) = 4$ . The value of $q$ such that the vector $\to a = p^i + q^j + r^k$ satisfies the relation $^j \times (^j \times \to a) = \to 0,$ is